Ex 5.7 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: The function \( \sin^{-1} x \) has the derivative \[ \frac{d}{dx}\sin^{-1} x = \frac{1}{\sqrt{1-x^2}}. \] This derivative is defined when the denominator is nonzero and real, i.e., when \[ 1 – x^2 > 0 \quad \Longrightarrow \quad -1 < x < 1. \]

Solution: The derivative of \( \cos^{-1} x \) is given by \[ \frac{d}{dx}\cos^{-1} x = -\frac{1}{\sqrt{1-x^2}}. \] This expression is defined when the denominator is real and nonzero, i.e., when \[ 1 – x^2 > 0 \quad \Longrightarrow \quad -1 < x < 1. \]

Solution: The derivative of \( \sec^{-1} x \) is given by \[ \frac{d}{dx} \sec^{-1} x = \frac{1}{|x|\sqrt{x^2-1}}. \] For the derivative to exist, the denominator must be defined and nonzero, which requires:

• \( \sqrt{x^2-1} \) is real, i.e., \( x^2-1 > 0 \) or \( |x| > 1 \).
• \( |x| \neq 0 \) is automatically satisfied when \( |x| > 1 \).

Thus, the function is differentiable when \( x < -1 \) or \( x > 1 \).

Solution: The derivative of \( \text{cosec}^{-1} x \) is given by \[ \frac{d}{dx}\text{cosec}^{-1} x = -\frac{1}{|x|\sqrt{x^2-1}}. \] For the derivative to be defined, the denominator \( |x|\sqrt{x^2-1} \) must be nonzero and real. This requires:

• \( \sqrt{x^2-1} \) is real, i.e., \( x^2-1 > 0 \) which implies \( |x| > 1 \).
• \( |x| \) is nonzero, which is automatically satisfied when \( |x| > 1 \).

Therefore, \( \text{cosec}^{-1} x \) is differentiable when \( x < -1 \) or \( x > 1 \).

Solution: The derivative of \( \tan^{-1} x \) is given by \[ \frac{d}{dx}\tan^{-1} x = \frac{1}{1+x^2}. \] Since \( 1+x^2 > 0 \) for all \( x \in \mathbb{R} \), the derivative is defined for all real numbers.

Solution: The derivative of \( \cot^{-1} x \) is given by \[ \frac{d}{dx}\cot^{-1} x = -\frac{1}{1+x^2}. \] Since \( 1+x^2 > 0 \) for all \( x \in \mathbb{R} \), the derivative is defined for every real number.

Solution: First, note that \[ \tan^{-1}(x\sqrt{x}) = \tan^{-1}\left(x^{\frac{3}{2}}\right). \] Differentiating using the chain rule, we have: \[ \frac{d}{dx}\tan^{-1}\left(x^{\frac{3}{2}}\right) = \frac{1}{1+\left(x^{\frac{3}{2}}\right)^2}\cdot \frac{d}{dx}\left(x^{\frac{3}{2}}\right). \] Since \[ \left(x^{\frac{3}{2}}\right)^2 = x^3 \quad \text{and} \quad \frac{d}{dx}\left(x^{\frac{3}{2}}\right) = \frac{3}{2}x^{\frac{1}{2}}, \] it follows that: \[ \frac{d}{dx}\tan^{-1}\left(x^{\frac{3}{2}}\right) = \frac{\frac{3}{2}x^{\frac{1}{2}}}{1+x^3}. \]

Solution: Let \( u = 3x \). Then, using the chain rule and the derivative formula for the inverse cosine function, we have: \[ \frac{d}{dx}\cos^{-1}(u) = -\frac{u’}{\sqrt{1-u^2}}. \] Here, \( u’ = 3 \) and \( u^2 = (3x)^2 = 9x^2 \). Therefore, \[ \frac{d}{dx}\cos^{-1}(3x) = -\frac{3}{\sqrt{1-9x^2}}. \]

Solution: Let \[ y = \sqrt{\sin^{-1}(2x)}. \] Using the chain rule, the derivative is: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin^{-1}(2x)}} \cdot \frac{d}{dx}[\sin^{-1}(2x)]. \] Now, \[ \frac{d}{dx}[\sin^{-1}(2x)] = \frac{2}{\sqrt{1 – (2x)^2}} = \frac{2}{\sqrt{1 – 4x^2}}. \] So, \[ \frac{dy}{dx} = \frac{1}{2\sqrt{\sin^{-1}(2x)}} \cdot \frac{2}{\sqrt{1 – 4x^2}} = \frac{1}{\sqrt{\sin^{-1}(2x)} \cdot \sqrt{1 – 4x^2}}. \]

Solution: Let \( u = \tan^{-1} x \) so that the function becomes \( \sin u \). Using the chain rule, we have: \[ \frac{d}{dx} \sin u = \cos u \cdot \frac{du}{dx}. \] Since \( \frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2} \), we get: \[ \frac{du}{dx} = \frac{1}{1+x^2}. \] Thus, the derivative is: \[ \frac{d}{dx}\sin\left(\tan^{-1} x\right) = \cos\left(\tan^{-1} x\right) \cdot \frac{1}{1+x^2}. \]

Solution: Let \( u = \sqrt{x} \). Then, the function becomes \( \cot^{-1} u \). We know that: \[ \frac{d}{du} \cot^{-1} u = -\frac{1}{1+u^2}. \] Also, the derivative of \( u = \sqrt{x} \) is: \[ \frac{du}{dx} = \frac{1}{2\sqrt{x}}. \] By the chain rule, we have: \[ \frac{d}{dx} \cot^{-1} \sqrt{x} = -\frac{1}{1+u^2} \cdot \frac{1}{2\sqrt{x}}. \] Since \( u^2 = x \), this simplifies to: \[ \frac{d}{dx} \cot^{-1} \sqrt{x} = -\frac{1}{2\sqrt{x}(1+x)}. \]

Solution: Let \( y = \left(\tan^{-1} x\right)^2 \). Differentiating using the chain rule, we get: \[ \frac{dy}{dx} = 2 \tan^{-1} x \cdot \frac{d}{dx}\left(\tan^{-1} x\right). \] Since: \[ \frac{d}{dx}\left(\tan^{-1} x\right) = \frac{1}{1+x^2}, \] it follows that: \[ \frac{dy}{dx} = \frac{2 \tan^{-1} x}{1+x^2}. \]

Solution: Let \[ y = \sec^{-1}(\text{cosec}\,x) \] We know that: \[ \text{cosec}\,x = \frac{1}{\sin x} \] So, \[ y = \sec^{-1}\left( \frac{1}{\sin x} \right) \] Now, we use the identity: \[ \sec^{-1}(u) = \cos^{-1}\left( \frac{1}{u} \right) \Rightarrow y = \cos^{-1}(\sin x) \] Now differentiate using the chain rule: \[ \frac{dy}{dx} = \frac{d}{dx} \cos^{-1}(\sin x) = -\frac{\cos x}{\sqrt{1 – \sin^2 x}} \] Simplify the denominator: \[ \sqrt{1 – \sin^2 x} = \sqrt{\cos^2 x} = |\cos x| \] So, \[ \frac{dy}{dx} = -\frac{\cos x}{|\cos x|} = -1 \quad \text{(for } \cos x > 0 \text{)} \]

Solution: Let \[ y = \tan^{-1}(\cot x). \] We know that \[ \cot x = \tan\left(\frac{\pi}{2}-x\right). \] Hence, \[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{2}-x\right)\right). \] For \(\frac{\pi}{2}-x\) within the principal range of \(\tan^{-1}\), we have: \[ \tan^{-1}\left(\tan\left(\frac{\pi}{2}-x\right)\right) = \frac{\pi}{2}-x. \] Differentiating with respect to \(x\) yields: \[ \frac{dy}{dx} = -1. \]

Solution: We differentiate \( f(x) = \cot^{-1}(\tan x) \) using the chain rule. First, recall that the derivative of \( \cot^{-1} u \) with respect to \( u \) is: \[ \frac{d}{du}\cot^{-1} u = -\frac{1}{1+u^2}. \] Let \( u = \tan x \). Then, we have: \[ \frac{du}{dx} = \sec^2 x. \] Applying the chain rule: \[ f'(x) = -\frac{1}{1+(\tan x)^2} \cdot \sec^2 x. \] Using the trigonometric identity \( 1+\tan^2 x = \sec^2 x \), we obtain: \[ f'(x) = -\frac{\sec^2 x}{\sec^2 x} = -1. \]

Solution: We differentiate each term separately. For the first term: \[ \frac{d}{dx}\sin^{-1} x = \frac{1}{\sqrt{1-x^2}}. \] For the second term, let \( u=\frac{1}{x} \) so that: \[ \frac{du}{dx} = -\frac{1}{x^2}. \] The derivative of \( \sec^{-1} u \) is: \[ \frac{d}{du}\sec^{-1} u = \frac{1}{|u|\sqrt{u^2-1}}. \] Substituting \( u=\frac{1}{x} \), we have: \[ |u|=\frac{1}{|x|} \quad \text{and} \quad u^2-1=\frac{1}{x^2}-1=\frac{1-x^2}{x^2}, \] so that: \[ \sqrt{u^2-1}=\frac{\sqrt{1-x^2}}{|x|}. \] Thus, the derivative of the second term is: \[ \frac{d}{dx}\sec^{-1}\left(\frac{1}{x}\right) = \frac{-\frac{1}{x^2}}{\frac{1}{|x|}\cdot\frac{\sqrt{1-x^2}}{|x|}} = -\frac{1}{\sqrt{1-x^2}}. \] Adding the two derivatives: \[ f'(x)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0. \]

Solution: We have: \[ y=\tan^{-1} x+\cot^{-1}\frac{1}{x}. \] Differentiating the first term: \[ \frac{d}{dx}\tan^{-1} x=\frac{1}{1+x^2}. \] For the second term, let \( u=\frac{1}{x} \) so that: \[ \frac{du}{dx}=-\frac{1}{x^2}, \] and \[ \frac{d}{du}\cot^{-1} u=-\frac{1}{1+u^2}. \] By the chain rule: \[ \frac{d}{dx}\cot^{-1}\frac{1}{x} = -\frac{1}{1+\left(\frac{1}{x}\right)^2}\cdot\left(-\frac{1}{x^2}\right) = \frac{1}{1+\frac{1}{x^2}}\cdot\frac{1}{x^2}. \] Notice that: \[ 1+\frac{1}{x^2}=\frac{x^2+1}{x^2}, \] so that: \[ \frac{1}{1+\frac{1}{x^2}}=\frac{x^2}{x^2+1}. \] Therefore: \[ \frac{d}{dx}\cot^{-1}\frac{1}{x}=\frac{x^2}{x^2+1}\cdot\frac{1}{x^2} =\frac{1}{1+x^2}. \] Adding the derivatives of both terms: \[ \frac{dy}{dx}=\frac{1}{1+x^2}+\frac{1}{1+x^2} =\frac{2}{1+x^2}. \]

Solution: We start by simplifying each term using trigonometric identities. For the first term, note that \[ \cos x = \sin\left(\frac{\pi}{2}-x\right). \] Hence, \[ \sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\frac{\pi}{2}-x\right)\right) = \frac{\pi}{2}-x, \] provided that \(\frac{\pi}{2}-x\) lies in the principal range of \(\sin^{-1}\). For the second term, observe that \[ \sec x = \frac{1}{\cos x} = \csc\left(\frac{\pi}{2}-x\right), \] so that \[ \csc^{-1}(\sec x) = \csc^{-1}\left(\csc\left(\frac{\pi}{2}-x\right)\right) = \frac{\pi}{2}-x, \] assuming the appropriate range for the inverse cosecant function. Therefore, the function becomes: \[ y=\left(\frac{\pi}{2}-x\right)+3\left(\frac{\pi}{2}-x\right)=4\left(\frac{\pi}{2}-x\right)=2\pi-4x. \] Differentiating with respect to \(x\) yields: \[ \frac{dy}{dx}=-4. \]

Solution: We differentiate using the product rule. Let: \[ u = x \quad \text{and} \quad v = \tan^{-1} x. \] Then, \[ u’ = 1 \quad \text{and} \quad v’ = \frac{1}{1+x^2}. \] Applying the product rule: \[ \frac{d}{dx}\left(x \tan^{-1} x\right) = u’v + uv’ = \tan^{-1} x + x \cdot \frac{1}{1+x^2}. \] Thus, the derivative is: \[ \tan^{-1} x + \frac{x}{1+x^2}. \]

Solution: We differentiate using the product rule by letting: \[ u = x \quad \text{and} \quad v = \cos^{-1} \sqrt{x}. \] Then, \( u’ = 1 \) and we differentiate \( v \) using the chain rule. Let \( w = \sqrt{x} \) so that \( v = \cos^{-1} w \) and \( \frac{dw}{dx} = \frac{1}{2\sqrt{x}} \). The derivative of \( \cos^{-1} w \) is: \[ \frac{dv}{dw} = -\frac{1}{\sqrt{1-w^2}}, \] thus, \[ v’ = -\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{2\sqrt{x}\sqrt{1-x}}. \] Now, applying the product rule: \[ \frac{d}{dx}\left( x \cos^{-1} \sqrt{x} \right) = u’v + uv’ = \cos^{-1} \sqrt{x} – \frac{x}{2\sqrt{x}\sqrt{1-x}}. \] Simplifying: \[ \frac{x}{2\sqrt{x}\sqrt{1-x}} = \frac{\sqrt{x}}{2\sqrt{1-x}}, \] so the derivative is: \[ \cos^{-1} \sqrt{x} – \frac{\sqrt{x}}{2\sqrt{1-x}}. \]

Solution: Let \( y = \frac{\sin^{-1} x}{x} \). Using the quotient rule where \[ u = \sin^{-1} x \quad \text{and} \quad v = x, \] we have: \[ u’ = \frac{1}{\sqrt{1-x^2}} \quad \text{and} \quad v’ = 1. \] The quotient rule gives: \[ y’=\frac{u’v-uv’}{v^2} = \frac{\frac{1}{\sqrt{1-x^2}}\cdot x-\sin^{-1} x\cdot 1}{x^2} = \frac{\frac{x}{\sqrt{1-x^2}}-\sin^{-1} x}{x^2}. \] Multiplying numerator and denominator by \(\sqrt{1-x^2}\), we get: \[ y’=\frac{x-\sin^{-1} x\,\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}. \]

Solution: Let \[ y = \cos^{-1}\left(\frac{1-x}{1+x}\right) \] and denote \[ u = \frac{1-x}{1+x}. \] Then, using the chain rule: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}. \] First, differentiate \( u \): \[ u = \frac{1-x}{1+x} \quad \Longrightarrow \quad \frac{du}{dx} = \frac{- (1+x) – (1-x)}{(1+x)^2} = -\frac{2}{(1+x)^2}. \] Next, compute \( \sqrt{1-u^2} \). We have: \[ u^2 = \left(\frac{1-x}{1+x}\right)^2 \quad \Longrightarrow \quad 1-u^2 = \frac{(1+x)^2-(1-x)^2}{(1+x)^2}. \] Note that: \[ (1+x)^2 – (1-x)^2 = \left(1+2x+x^2\right)-\left(1-2x+x^2\right) = 4x, \] so that: \[ 1-u^2 = \frac{4x}{(1+x)^2} \quad \Longrightarrow \quad \sqrt{1-u^2} = \frac{2\sqrt{x}}{1+x}. \] Substituting these into the derivative: \[ \frac{dy}{dx} = -\frac{1}{\frac{2\sqrt{x}}{1+x}} \cdot \left(-\frac{2}{(1+x)^2}\right) = \frac{2}{(1+x)^2} \cdot \frac{1+x}{2\sqrt{x}} = \frac{1}{(1+x)\sqrt{x}}. \]

Solution: Let \[ y = \sec^{-1}\left(\frac{1}{x-1}\right), \] and define \[ u = \frac{1}{x-1}. \] The derivative of \( \sec^{-1} u \) with respect to \( u \) is given by: \[ \frac{d}{du} \left(\sec^{-1} u\right) = \frac{1}{|u|\sqrt{u^2-1}}. \] First, differentiate \( u \) with respect to \( x \): \[ u = \frac{1}{x-1} \quad \Longrightarrow \quad u’ = -\frac{1}{(x-1)^2}. \] Next, note that: \[ u^2 = \frac{1}{(x-1)^2} \quad \Longrightarrow \quad u^2-1 = \frac{1-(x-1)^2}{(x-1)^2}. \] Also, \( |u| = \frac{1}{|x-1|} \). Now, applying the chain rule: \[ \frac{dy}{dx} = \frac{u’}{|u|\sqrt{u^2-1}} = \frac{-\frac{1}{(x-1)^2}}{\frac{1}{|x-1|}\sqrt{\frac{1-(x-1)^2}{(x-1)^2}}}. \] Simplify the square root: \[ \sqrt{\frac{1-(x-1)^2}{(x-1)^2}} = \frac{\sqrt{1-(x-1)^2}}{|x-1|}. \] Substituting back, we have: \[ \frac{dy}{dx} = -\frac{1}{(x-1)^2} \cdot \frac{|x-1|}{\frac{\sqrt{1-(x-1)^2}}{|x-1|}} = -\frac{1}{(x-1)^2} \cdot \frac{|x-1|^2}{\sqrt{1-(x-1)^2}}. \] Since \( |x-1|^2=(x-1)^2 \), the expression simplifies to: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-(x-1)^2}}. \] Finally, note that: \[ 1-(x-1)^2 = 1-(x^2-2x+1)=2x-x^2. \] Therefore, the derivative is: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{2x-x^2}}. \]

Solution: Let \[ y = \sin^{-1}\left(\frac{1}{\sqrt{x+1}}\right). \] Define \[ u = \frac{1}{\sqrt{x+1}} = (x+1)^{-1/2}. \] Then, by the chain rule, \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}. \] First, compute \( \frac{du}{dx} \): \[ \frac{du}{dx} = -\frac{1}{2}(x+1)^{-3/2}. \] Next, note that \[ u^2 = \frac{1}{x+1}, \quad \text{so} \quad 1-u^2 = 1 – \frac{1}{x+1} = \frac{x}{x+1}. \] Hence, \[ \sqrt{1-u^2} = \sqrt{\frac{x}{x+1}} = \frac{\sqrt{x}}{\sqrt{x+1}}. \] Substituting these into the derivative, \[ \frac{dy}{dx} = \frac{-\frac{1}{2}(x+1)^{-3/2}}{\frac{\sqrt{x}}{\sqrt{x+1}}} = -\frac{1}{2} (x+1)^{-3/2} \cdot \frac{\sqrt{x+1}}{\sqrt{x}} = -\frac{1}{2} \frac{1}{(x+1)\sqrt{x}}. \]

Solution: Let \[ y = \sin^{-1} x + \sin^{-1}\sqrt{1-x^2}. \] Set \( u = \sin^{-1} x \). Then, by the Pythagorean identity, \[ \sqrt{1-x^2} = \cos u. \] Hence, \[ \sin^{-1}\sqrt{1-x^2} = \sin^{-1}(\cos u). \] For \( u \) in the range \([0, \frac{\pi}{2}]\), we have the identity: \[ \sin^{-1}(\cos u) = \frac{\pi}{2} – u. \] Therefore, \[ y = u + \left(\frac{\pi}{2} – u\right) = \frac{\pi}{2}. \] Since \( y \) is a constant, its derivative is: \[ \frac{dy}{dx} = 0. \]

Solution: Let \[ u=\sin^{-1}x. \] Then the given function can be written as: \[ y=x\,u^2+2\sqrt{1-x^2}\,u-2x. \] We differentiate term by term. \(\mathbf{Term\ 1:}\) For \(x\,u^2\), using the product rule: \[ \frac{d}{dx}\left(x\,u^2\right)=u^2+x\cdot2u\cdot\frac{du}{dx}. \] Since \[ \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}, \] this becomes: \[ u^2+\frac{2xu}{\sqrt{1-x^2}}. \] \(\mathbf{Term\ 2:}\) For \(2\sqrt{1-x^2}\,u\), again applying the product rule: \[ \frac{d}{dx}\left(2\sqrt{1-x^2}\,u\right)=2\left[\frac{d}{dx}\left(\sqrt{1-x^2}\right)u+\sqrt{1-x^2}\cdot\frac{du}{dx}\right]. \] We know: \[ \frac{d}{dx}\left(\sqrt{1-x^2}\right)=\frac{-x}{\sqrt{1-x^2}}, \] so: \[ 2\left[-\frac{xu}{\sqrt{1-x^2}}+\sqrt{1-x^2}\cdot\frac{1}{\sqrt{1-x^2}}\right]=2\left[-\frac{xu}{\sqrt{1-x^2}}+1\right]. \] \(\mathbf{Term\ 3:}\) The derivative of \(-2x\) is \(-2\). Now, summing the derivatives of all terms: \[ y’=\left[u^2+\frac{2xu}{\sqrt{1-x^2}}\right]+2\left(1-\frac{xu}{\sqrt{1-x^2}}\right)-2. \] Notice that: \[ \frac{2xu}{\sqrt{1-x^2}}-2\frac{xu}{\sqrt{1-x^2}}=0, \] and: \[ 2-2=0. \] Therefore, the derivative simplifies to: \[ y’=u^2=\left(\sin^{-1}x\right)^2. \]

Solution: Let \[ y = \tan^{-1}(\sin x+\cos x). \] Using the chain rule, we have: \[ \frac{dy}{dx} = \frac{1}{1+(\sin x+\cos x)^2}\cdot\frac{d}{dx}(\sin x+\cos x). \] Differentiating \(\sin x+\cos x\) with respect to \(x\): \[ \frac{d}{dx}(\sin x+\cos x) = \cos x-\sin x. \] Next, simplify the term \((\sin x+\cos x)^2\): \[ (\sin x+\cos x)^2 = \sin^2x+2\sin x\cos x+\cos^2x = 1+ \sin 2x. \] Thus, the derivative becomes: \[ \frac{dy}{dx} = \frac{\cos x-\sin x}{1+(1+\sin2x)} = \frac{\cos x-\sin x}{2+\sin2x}. \]

Solution: Let \[ y = \tan^{-1}\left(\frac{\sin x}{1+\cos x}\right). \] We know that using the half-angle identity: \[ \tan\frac{x}{2} = \frac{\sin x}{1+\cos x}. \] Therefore, \[ y = \tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}, \] provided that \(\frac{x}{2}\) lies within the principal range of \(\tan^{-1}\). Differentiating with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{2}. \]

Solution: Let \[ y = \tan^{-1}\left(\frac{1+\cos x}{\sin x}\right). \] Using the half-angle identity, we know that \[ \cot\frac{x}{2} = \frac{1+\cos x}{\sin x}. \] Thus, we can write: \[ y = \tan^{-1}\left(\cot\frac{x}{2}\right). \] Since \[ \cot\frac{x}{2} = \tan\left(\frac{\pi}{2}-\frac{x}{2}\right), \] it follows that \[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{2}-\frac{x}{2}\right)\right)= \frac{\pi}{2}-\frac{x}{2}, \] assuming the angle \(\frac{\pi}{2}-\frac{x}{2}\) lies in the principal range of \(\tan^{-1}\). Differentiating with respect to \(x\): \[ \frac{dy}{dx} = -\frac{1}{2}. \]

Solution: Let \[ y = \tan^{-1}\left(\frac{\cos x}{1+\sin x}\right). \] We use the half-angle identity: \[ \cot\frac{x}{2} = \frac{\cos x}{1+\sin x}. \] Thus, \[ y = \tan^{-1}\left(\cot\frac{x}{2}\right). \] Since \[ \cot\frac{x}{2} = \tan\left(\frac{\pi}{2}-\frac{x}{2}\right), \] we have: \[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{2}-\frac{x}{2}\right)\right) = \frac{\pi}{2}-\frac{x}{2}, \] assuming the angle lies within the principal range of \(\tan^{-1}\). Differentiating with respect to \(x\): \[ \frac{dy}{dx} = -\frac{1}{2}. \]

Solution: Let \[ y = \cot^{-1}(\csc x+\cot x). \] We use the trigonometric identity: \[ \csc x+\cot x = \cot\frac{x}{2}. \] Hence, \[ y = \cot^{-1}\left(\cot\frac{x}{2}\right). \] Since the principal value of \(\cot^{-1}\) lies in \((0,\pi)\) and \(\frac{x}{2}\) is assumed to be in this range, we have: \[ \cot^{-1}\left(\cot\frac{x}{2}\right)=\frac{x}{2}. \] Differentiating with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{2}. \]

Solution: Let \[ y = \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right). \] Recognize the half-angle identity: \[ \tan\frac{x}{2} = \sqrt{\frac{1-\cos x}{1+\cos x}}. \] Therefore, \[ y = \tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}, \] provided that \(\frac{x}{2}\) lies within the principal range of \(\tan^{-1}\). Differentiating with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{2}. \]

Solution: Let \[ y = \cot^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right). \] We use the identity: \[ \tan\left(\frac{\pi}{4}+\frac{x}{2}\right) = \sqrt{\frac{1+\sin x}{1-\sin x}}, \] which implies that \[ \sqrt{\frac{1-\sin x}{1+\sin x}} = \cot\left(\frac{\pi}{4}+\frac{x}{2}\right). \] Hence, we can rewrite \( y \) as: \[ y = \cot^{-1}\left(\cot\left(\frac{\pi}{4}+\frac{x}{2}\right)\right). \] Since the inverse cotangent of the cotangent of an angle returns that angle (assuming the angle lies within the principal range), we have: \[ y = \frac{\pi}{4}+\frac{x}{2}. \] Differentiating with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}+\frac{x}{2}\right) = \frac{1}{2}. \]

Solution: Let \[ f(x)=\tan^{-1}\left(\frac{2x}{1-x^2}\right) \] and define \[ u(x)=\frac{2x}{1-x^2}. \] Then, using the chain rule, \[ f'(x)=\frac{u'(x)}{1+u(x)^2}. \] First, differentiate \( u(x) \) using the quotient rule: \[ u'(x)=\frac{(2)(1-x^2)-2x(-2x)}{(1-x^2)^2} = \frac{2(1-x^2)+4x^2}{(1-x^2)^2} = \frac{2(1+x^2)}{(1-x^2)^2}. \] Next, compute \( u(x)^2 \): \[ u(x)^2=\left(\frac{2x}{1-x^2}\right)^2=\frac{4x^2}{(1-x^2)^2}. \] Therefore, \[ 1+u(x)^2 = \frac{(1-x^2)^2+4x^2}{(1-x^2)^2} = \frac{1-2x^2+x^4+4x^2}{(1-x^2)^2} = \frac{1+2x^2+x^4}{(1-x^2)^2} = \frac{(1+x^2)^2}{(1-x^2)^2}. \] Now, substitute \( u'(x) \) and \( 1+u(x)^2 \) into the derivative: \[ f'(x)=\frac{\frac{2(1+x^2)}{(1-x^2)^2}}{\frac{(1+x^2)^2}{(1-x^2)^2}} =\frac{2(1+x^2)}{(1+x^2)^2} =\frac{2}{1+x^2}. \]

Solution: Let \[ f(x)=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \] and set \[ u(x)=\frac{1-x^2}{1+x^2}. \] Then, by the chain rule, \[ f'(x)=-\frac{u'(x)}{\sqrt{1-u(x)^2}}. \] First, differentiate \( u(x) \) using the quotient rule: \[ u'(x)=\frac{(-2x)(1+x^2) – (1-x^2)(2x)}{(1+x^2)^2} = \frac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2)^2}. \] Notice that: \[ -2x(1+x^2)-2x(1-x^2) = -2x\Big[(1+x^2)+(1-x^2)\Big] = -2x(2)= -4x. \] Hence, \[ u'(x)=\frac{-4x}{(1+x^2)^2}. \] Next, compute \( 1-u(x)^2 \): \[ u(x)^2=\left(\frac{1-x^2}{1+x^2}\right)^2=\frac{(1-x^2)^2}{(1+x^2)^2}, \] so that \[ 1-u(x)^2=\frac{(1+x^2)^2-(1-x^2)^2}{(1+x^2)^2}. \] Expanding the numerator: \[ (1+x^2)^2-(1-x^2)^2=\left(1+2x^2+x^4\right)-\left(1-2x^2+x^4\right)=4x^2. \] Therefore, \[ 1-u(x)^2=\frac{4x^2}{(1+x^2)^2} \quad \text{and} \quad \sqrt{1-u(x)^2}=\frac{2|x|}{1+x^2}. \] Now, substituting \( u'(x) \) and \( \sqrt{1-u(x)^2} \) into the derivative: \[ f'(x)=-\frac{-4x/(1+x^2)^2}{2|x|/(1+x^2)} =\frac{4x}{(1+x^2)^2}\cdot\frac{1+x^2}{2|x|} =\frac{4x}{2|x|(1+x^2)} =\frac{2x}{|x|(1+x^2)}. \] Recognizing that \[ \frac{x}{|x|}=\text{sgn}(x), \] we have: \[ f'(x)=\frac{2\,\text{sgn}(x)}{1+x^2}. \] For \( x>0 \), this reduces to: \[ f'(x)=\frac{2}{1+x^2}. \]

Solution: Let \[ f(x)=\cos^{-1}\left(1-2x^2\right). \] Using the chain rule, the derivative is given by: \[ f'(x)=-\frac{d}{dx}\left(1-2x^2\right)\Big/\sqrt{1-\left(1-2x^2\right)^2}. \] First, differentiate the inner function: \[ \frac{d}{dx}\left(1-2x^2\right)=-4x. \] Therefore, \[ f'(x)=-\frac{-4x}{\sqrt{1-\left(1-2x^2\right)^2}} =\frac{4x}{\sqrt{1-\left(1-2x^2\right)^2}}. \] Next, simplify the expression under the square root: \[ 1-\left(1-2x^2\right)^2 = 1-\left(1-4x^2+4x^4\right) = 4x^2-4x^4 = 4x^2\left(1-x^2\right). \] Hence, \[ \sqrt{1-\left(1-2x^2\right)^2} = \sqrt{4x^2\left(1-x^2\right)} =2|x|\sqrt{1-x^2}. \] Substituting back into the derivative: \[ f'(x)=\frac{4x}{2|x|\sqrt{1-x^2}} =\frac{2x}{|x|\sqrt{1-x^2}}. \] Notice that \[ \frac{x}{|x|}=\text{sgn}(x), \] where \(\text{sgn}(x)\) is the sign function. Thus, \[ f'(x)=\frac{2\,\text{sgn}(x)}{\sqrt{1-x^2}}. \] In particular, if \( x>0 \) then \(\text{sgn}(x)=1\) and \[ f'(x)=\frac{2}{\sqrt{1-x^2}}. \]

Solution: Let \[ f(x)=\sin^{-1}\left(2x\sqrt{1-x^2}\right) \] and define \[ u(x)=2x\sqrt{1-x^2}. \] The derivative of \( f(x) \) is given by: \[ f'(x)=\frac{u'(x)}{\sqrt{1-u(x)^2}}. \] First, we compute \( u(x)^2 \): \[ u(x)^2=4x^2(1-x^2)=4x^2-4x^4. \] Therefore, \[ 1-u(x)^2=1-4x^2+4x^4=(1-2x^2)^2, \] so that \[ \sqrt{1-u(x)^2}=|1-2x^2|. \] Next, differentiate \( u(x)=2x\sqrt{1-x^2} \) using the product rule: \[ u'(x)=2\sqrt{1-x^2}+2x\cdot\frac{-x}{\sqrt{1-x^2}} =2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}. \] Writing over a common denominator, we have: \[ u'(x)=\frac{2(1-x^2)-2x^2}{\sqrt{1-x^2}} =\frac{2-4x^2}{\sqrt{1-x^2}} =\frac{2(1-2x^2)}{\sqrt{1-x^2}}. \] Now, substituting \( u'(x) \) and \( \sqrt{1-u(x)^2} \) into the derivative formula gives: \[ f'(x)=\frac{2(1-2x^2)/\sqrt{1-x^2}}{|1-2x^2|} =\frac{2}{\sqrt{1-x^2}}\cdot\frac{1-2x^2}{|1-2x^2|}. \] The fraction \(\frac{1-2x^2}{|1-2x^2|}\) represents the sign of \(1-2x^2\). Hence, we can write: \[ f'(x)=\frac{2\,\text{sgn}(1-2x^2)}{\sqrt{1-x^2}}, \] where \[ \text{sgn}(1-2x^2)= \begin{cases} 1, & \text{if } 1-2x^2>0,\\[4pt] -1, & \text{if } 1-2x^2<0. \end{cases} \] For the principal branch (typically when \(1-2x^2>0\)), this simplifies to: \[ f'(x)=\frac{2}{\sqrt{1-x^2}}. \]

Solution: Notice that if we set \[ x=\sin \theta, \] then we have the trigonometric identity: \[ 3\sin\theta-4\sin^3\theta=\sin 3\theta. \] Hence, \[ f(x)=\sin^{-1}\left(3x-4x^3\right)=\sin^{-1}(\sin 3\theta). \] For values of \(\theta\) in the appropriate interval, this simplifies to: \[ f(x)=3\theta. \] Since \(x=\sin \theta\), we have: \[ \theta=\sin^{-1} x. \] Therefore, \[ f(x)=3\sin^{-1} x. \] Differentiating with respect to \(x\): \[ f'(x)=3\cdot\frac{d}{dx}\left(\sin^{-1} x\right) =\frac{3}{\sqrt{1-x^2}}. \]

Solution: Let \[ f(x)=\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right). \] Substitute \[ x=\tan \theta. \] Then, the expression inside the arctan becomes \[ \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}. \] Recall the triple angle formula for tangent: \[ \tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}. \] Hence, \[ f(x)=\tan^{-1}(\tan 3\theta). \] For values of \(\theta\) where the inverse tangent and tangent functions cancel appropriately, we have: \[ f(x)=3\theta. \] Since \(x=\tan \theta\), it follows that: \[ \theta=\tan^{-1}x. \] Therefore, the function can be written as: \[ f(x)=3\tan^{-1}x. \] Differentiating with respect to \(x\): \[ f'(x)=3\cdot\frac{d}{dx}\left(\tan^{-1}x\right) =\frac{3}{1+x^2}. \]

Solution: Let \[ f(x)=\cos^{-1}\left(\frac{2x}{1+x^2}\right). \] Substitute \( x=\tan\theta \). Then, note that: \[ \frac{2x}{1+x^2}=\frac{2\tan\theta}{1+\tan^2\theta}=\sin 2\theta. \] Hence, we have: \[ f(x)=\cos^{-1}(\sin 2\theta). \] Using the identity \(\sin 2\theta=\cos\Big(\frac{\pi}{2}-2\theta\Big)\), it follows that: \[ f(x)=\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-2\theta\Big)\Big) =\frac{\pi}{2}-2\theta, \] provided the angle is in the appropriate range. Since \( \theta=\tan^{-1}x \), we obtain: \[ f(x)=\frac{\pi}{2}-2\tan^{-1}x. \] Differentiating with respect to \(x\): \[ f'(x)=-2\cdot\frac{d}{dx}\Big(\tan^{-1}x\Big) =-\frac{2}{1+x^2}. \]

Solution: To simplify the expression, set \[ x=\cos \theta. \] Then, notice that: \[ 4x^3-3x=4\cos^3\theta-3\cos\theta=\cos3\theta. \] Hence, \[ \frac{1}{4x^3-3x}=\frac{1}{\cos3\theta}=\sec3\theta. \] Therefore, \[ f(x)=\sec^{-1}\left(\frac{1}{4x^3-3x}\right)=\sec^{-1}(\sec3\theta). \] For the principal value, we have: \[ f(x)=3\theta. \] Since \(x=\cos\theta\), it follows that: \[ \theta=\cos^{-1}x. \] Thus, the function becomes: \[ f(x)=3\cos^{-1}x. \] Differentiating with respect to \(x\): \[ f'(x)=3\cdot\frac{d}{dx}\Big(\cos^{-1}x\Big) =3\left(-\frac{1}{\sqrt{1-x^2}}\right) =-\frac{3}{\sqrt{1-x^2}}. \]

Solution: Let \[ y=\sec^{-1}\left(\frac{x^2+1}{x^2-1}\right) \] and set \[ u=\frac{x^2+1}{x^2-1}. \] Then \( y=\sec^{-1}(u) \) and by the chain rule, the derivative is \[ y’=\frac{u’}{|u|\sqrt{u^2-1}}. \] First, differentiate \( u \) using the quotient rule: \[ u’=\frac{d}{dx}\left(\frac{x^2+1}{x^2-1}\right) =\frac{(2x)(x^2-1)- (2x)(x^2+1)}{(x^2-1)^2} =\frac{-4x}{(x^2-1)^2}. \] Next, compute \( u^2-1 \): \[ u^2-1=\left(\frac{x^2+1}{x^2-1}\right)^2-1 =\frac{(x^2+1)^2-(x^2-1)^2}{(x^2-1)^2}. \] Notice that \[ (x^2+1)^2-(x^2-1)^2= \left[(x^2+1)-(x^2-1)\right]\left[(x^2+1)+(x^2-1)\right]=(2)(2x^2)=4x^2. \] Therefore, \[ u^2-1=\frac{4x^2}{(x^2-1)^2} \quad \text{and} \quad \sqrt{u^2-1}=\frac{2|x|}{|x^2-1|}. \] Substituting \( u’ \) and \( \sqrt{u^2-1} \) into the derivative formula: \[ y’=\frac{-4x}{(x^2-1)^2}\cdot\frac{|x^2-1|^2}{2|x|(x^2+1)} =-\frac{2x}{|x|(x^2+1)}. \] Since \( \frac{x}{|x|}=1 \) for \( x>0 \) (as per the typical domain considerations for \(\sec^{-1}\) functions), we have \[ y’=-\frac{2}{1+x^2}. \]

Solution: Let \[ y=\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) \] and denote \[ u=\frac{1-x^2}{1+x^2}. \] Then, by the chain rule, we have \[ y’=\frac{1}{\sqrt{1-u^2}}\, u’. \] First, differentiate \( u \) using the quotient rule: \[ u’=\frac{(1+x^2)\cdot(-2x) – (1-x^2)\cdot(2x)}{(1+x^2)^2} =\frac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2)^2}. \] Notice that \[ -2x(1+x^2)-2x(1-x^2)=-2x\Big[(1+x^2)+(1-x^2)\Big] =-2x(2)= -4x. \] Hence, \[ u’=\frac{-4x}{(1+x^2)^2}. \] Next, calculate \( 1-u^2 \): \[ 1-u^2=1-\left(\frac{1-x^2}{1+x^2}\right)^2 =\frac{(1+x^2)^2-(1-x^2)^2}{(1+x^2)^2}. \] Expanding, \[ (1+x^2)^2-(1-x^2)^2 = \bigl[1+2x^2+x^4\bigr]-\bigl[1-2x^2+x^4\bigr]=4x^2. \] Thus, \[ 1-u^2=\frac{4x^2}{(1+x^2)^2} \quad \text{and} \quad \sqrt{1-u^2}=\frac{2|x|}{1+x^2}. \] Substituting \( u’ \) and \( \sqrt{1-u^2} \) into the derivative: \[ y’=\frac{-4x}{(1+x^2)^2}\cdot\frac{1+x^2}{2|x|} = -\frac{4x}{2|x|(1+x^2)} = -\frac{2x}{|x|(1+x^2)}. \] For \( x>0 \), we have \( \frac{x}{|x|}=1 \), so the derivative simplifies to: \[ y’=-\frac{2}{1+x^2}. \]

Solution: We start with \[ y=\tan^{-1}\Bigl(\sqrt{1+x^2}-x\Bigr). \] A useful observation is that the expression inside the inverse tangent can be rewritten as \[ \sqrt{1+x^2}-x=\frac{1}{\sqrt{1+x^2}+x}, \] since \[ \left(\sqrt{1+x^2}-x\right)\left(\sqrt{1+x^2}+x\right)=1+x^2-x^2=1. \] Hence, we have \[ y=\tan^{-1}\left(\frac{1}{\sqrt{1+x^2}+x}\right). \] It is known that \[ \tan^{-1}\left(\sqrt{1+x^2}-x\right)=\frac{\pi}{4}-\frac{1}{2}\tan^{-1}x, \] which can be verified by employing the substitution \(x=\tan t\). Differentiating this equivalent form with respect to \(x\) yields: \[ y’=\frac{d}{dx}\left(\frac{\pi}{4}-\frac{1}{2}\tan^{-1}x\right) = -\frac{1}{2}\cdot\frac{1}{1+x^2}. \] Thus, the derivative is \[ y’=-\frac{1}{2(1+x^2)}. \]

Solution: Let \[ y=\cot^{-1}\left(\sqrt{1+x^2}+x\right). \] We know that \[ \cot^{-1}(u)=\tan^{-1}\left(\frac{1}{u}\right) \] for positive \( u \). Notice that \[ \sqrt{1+x^2}+x \quad \text{and} \quad \sqrt{1+x^2}-x \] are reciprocals since \[ \left(\sqrt{1+x^2}+x\right)\left(\sqrt{1+x^2}-x\right)=1. \] Hence, we can write: \[ y=\cot^{-1}\left(\sqrt{1+x^2}+x\right) =\tan^{-1}\left(\sqrt{1+x^2}-x\right). \] From part 14.(iii), we already have: \[ \frac{d}{dx}\tan^{-1}\left(\sqrt{1+x^2}-x\right) = -\frac{1}{2(1+x^2)}. \] Therefore, the derivative of \( y \) is: \[ y’=-\frac{1}{2(1+x^2)}. \]

Solution: Let \[ y=\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right). \] Define \[ u=\frac{x}{\sqrt{a^2-x^2}}. \] Then, by the chain rule, \[ y’=\frac{u’}{1+u^2}. \] First, rewrite \[ u=x(a^2-x^2)^{-1/2}. \] Differentiate \( u \) using the product rule: \[ u’=\frac{d}{dx}\left[x\right](a^2-x^2)^{-1/2} +x\frac{d}{dx}\left[(a^2-x^2)^{-1/2}\right]. \] Since \(\frac{d}{dx}x=1\) and \[ \frac{d}{dx}(a^2-x^2)^{-1/2} = -\frac{1}{2}(a^2-x^2)^{-3/2}(-2x) = \frac{x}{(a^2-x^2)^{3/2}}, \] we obtain: \[ u’=(a^2-x^2)^{-1/2}+ \frac{x^2}{(a^2-x^2)^{3/2}} =\frac{(a^2-x^2)+x^2}{(a^2-x^2)^{3/2}} =\frac{a^2}{(a^2-x^2)^{3/2}}. \] Next, calculate \( 1+u^2 \): \[ 1+u^2=1+\frac{x^2}{a^2-x^2}=\frac{(a^2-x^2)+x^2}{a^2-x^2} =\frac{a^2}{a^2-x^2}. \] Thus, the derivative is: \[ y’=\frac{\frac{a^2}{(a^2-x^2)^{3/2}}}{\frac{a^2}{a^2-x^2}} =\frac{a^2}{(a^2-x^2)^{3/2}}\cdot\frac{a^2-x^2}{a^2} =\frac{1}{\sqrt{a^2-x^2}}. \]

Solution: Let \[ y=\sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right). \] Set \[ u=\frac{1}{\sqrt{1+x^2}}=(1+x^2)^{-1/2}. \] Then, by the chain rule, \[ y’=\frac{u’}{\sqrt{1-u^2}}. \] First, differentiate \( u \): \[ u’=-\frac{1}{2}(1+x^2)^{-3/2}\cdot 2x =-\frac{x}{(1+x^2)^{3/2}}. \] Next, compute \( 1-u^2 \): \[ 1-u^2=1-\frac{1}{1+x^2}=\frac{(1+x^2)-1}{1+x^2} =\frac{x^2}{1+x^2}. \] Thus, \[ \sqrt{1-u^2}=\frac{|x|}{\sqrt{1+x^2}}. \] Assuming \( x>0 \) (so that \( |x|=x \)), we get: \[ \sqrt{1-u^2}=\frac{x}{\sqrt{1+x^2}}. \] Therefore, the derivative is: \[ y’=\frac{-\frac{x}{(1+x^2)^{3/2}}}{\frac{x}{\sqrt{1+x^2}}} = -\frac{1}{1+x^2}. \]

Solution: Let \[ y=\cos^{-1}\left(\sqrt{\frac{1+x}{2}}\right) \] and set \[ u=\sqrt{\frac{1+x}{2}}. \] Then, \( y=\cos^{-1}(u) \) and by the chain rule, \[ y’=-\frac{u’}{\sqrt{1-u^2}}. \] First, express \( u \) as \[ u=\left(\frac{1+x}{2}\right)^{\frac{1}{2}}. \] Differentiating with respect to \( x \) gives \[ u’=\frac{1}{2}\left(\frac{1+x}{2}\right)^{-\frac{1}{2}}\cdot\frac{1}{2} =\frac{1}{4}\left(\frac{2}{1+x}\right)^{\frac{1}{2}} =\frac{\sqrt{2}}{4\sqrt{1+x}}. \] Next, compute \( 1-u^2 \). Since \[ u^2=\frac{1+x}{2}, \] we have \[ 1-u^2=1-\frac{1+x}{2}=\frac{1-x}{2}. \] Therefore, \[ \sqrt{1-u^2}=\sqrt{\frac{1-x}{2}}=\frac{\sqrt{1-x}}{\sqrt{2}}. \] Substituting these into the derivative formula yields: \[ y’=-\frac{\frac{\sqrt{2}}{4\sqrt{1+x}}}{\frac{\sqrt{1-x}}{\sqrt{2}}} = -\frac{\sqrt{2}}{4\sqrt{1+x}}\cdot\frac{\sqrt{2}}{\sqrt{1-x}} = -\frac{2}{4\sqrt{1+x}\sqrt{1-x}} = -\frac{1}{2\sqrt{(1+x)(1-x)}}. \] Since \(\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}\), we have: \[ y’=-\frac{1}{2\sqrt{1-x^2}}. \]

Solution: Let \[ y=\sin^{-1}\left(\frac{x}{\sqrt{x^2+a^2}}\right). \] Define \[ u=\frac{x}{\sqrt{x^2+a^2}}. \] Then, by the chain rule, \[ y’=\frac{u’}{\sqrt{1-u^2}}. \] First, express \( u \) as \[ u=x\,(x^2+a^2)^{-1/2}. \] Differentiating using the product rule: \[ u’=\frac{d}{dx}\left(x\right)(x^2+a^2)^{-1/2}+x\frac{d}{dx}\left((x^2+a^2)^{-1/2}\right). \] Since \[ \frac{d}{dx}x=1, \] and \[ \frac{d}{dx}(x^2+a^2)^{-1/2}=-\frac{1}{2}(x^2+a^2)^{-3/2}\cdot 2x=-\frac{x}{(x^2+a^2)^{3/2}}, \] we obtain: \[ u’=(x^2+a^2)^{-1/2}-\frac{x^2}{(x^2+a^2)^{3/2}} =\frac{(x^2+a^2)-x^2}{(x^2+a^2)^{3/2}} =\frac{a^2}{(x^2+a^2)^{3/2}}. \] Next, compute \( \sqrt{1-u^2} \). Note that: \[ u^2=\frac{x^2}{x^2+a^2} \quad \Rightarrow \quad 1-u^2=1-\frac{x^2}{x^2+a^2}=\frac{a^2}{x^2+a^2}. \] Thus, \[ \sqrt{1-u^2}=\frac{a}{\sqrt{x^2+a^2}}. \] Substituting \( u’ \) and \( \sqrt{1-u^2} \) in the derivative formula: \[ y’=\frac{\frac{a^2}{(x^2+a^2)^{3/2}}}{\frac{a}{\sqrt{x^2+a^2}}} =\frac{a^2}{(x^2+a^2)^{3/2}} \cdot \frac{\sqrt{x^2+a^2}}{a} =\frac{a}{x^2+a^2}. \]

Solution: We start with \[ y=\tan^{-1}\sqrt{\frac{a-x}{a+x}}. \] Using the hint, let \[ x=a\cos t. \] Then, \[ \frac{a-x}{a+x}=\frac{a-a\cos t}{a+a\cos t}=\frac{1-\cos t}{1+\cos t}. \] Recall that \[ \tan^2\left(\frac{t}{2}\right)=\frac{1-\cos t}{1+\cos t}, \] so that \[ \sqrt{\frac{1-\cos t}{1+\cos t}}=\tan\left(\frac{t}{2}\right). \] Hence, we have \[ y=\tan^{-1}\left(\tan\left(\frac{t}{2}\right)\right)=\frac{t}{2}. \] Since \[ t=\cos^{-1}\left(\frac{x}{a}\right), \] it follows that \[ y=\frac{1}{2}\cos^{-1}\left(\frac{x}{a}\right). \] Differentiating with respect to \( x \): \[ y’=\frac{1}{2}\cdot\frac{d}{dx}\left[\cos^{-1}\left(\frac{x}{a}\right)\right] =\frac{1}{2}\cdot\left(-\frac{1}{\sqrt{1-\left(\frac{x}{a}\right)^2}}\cdot\frac{1}{a}\right) =-\frac{1}{2a\sqrt{1-\frac{x^2}{a^2}}}. \] Simplify further using \[ \sqrt{1-\frac{x^2}{a^2}}=\frac{\sqrt{a^2-x^2}}{a}, \] so that \[ y’=-\frac{1}{2a\cdot \frac{\sqrt{a^2-x^2}}{a}} = -\frac{1}{2\sqrt{a^2-x^2}}. \]

Solution: Let \[ y=\sin^{-1}\left(\sqrt{\frac{1+x^2}{2}}\right). \] Define \[ u=\sqrt{\frac{1+x^2}{2}}, \] so that \( y=\sin^{-1}(u) \) and by the chain rule: \[ y’=\frac{u’}{\sqrt{1-u^2}}. \] First, express \( u \) as \[ u=\left(\frac{1+x^2}{2}\right)^{1/2}. \] Differentiating with respect to \( x \): \[ u’=\frac{1}{2}\left(\frac{1+x^2}{2}\right)^{-1/2}\cdot\frac{d}{dx}\left(\frac{1+x^2}{2}\right) =\frac{1}{2}\left(\frac{1+x^2}{2}\right)^{-1/2}\cdot\frac{2x}{2} =\frac{x}{2\sqrt{\frac{1+x^2}{2}}}. \] Simplify: \[ u’=\frac{x}{2}\sqrt{\frac{2}{1+x^2}} =\frac{x}{\sqrt{2(1+x^2)}}. \] Next, compute: \[ u^2=\frac{1+x^2}{2} \quad \Rightarrow \quad 1-u^2=1-\frac{1+x^2}{2}=\frac{1-x^2}{2}. \] Thus, \[ \sqrt{1-u^2}=\sqrt{\frac{1-x^2}{2}}=\frac{\sqrt{1-x^2}}{\sqrt{2}}. \] Substituting into the derivative formula: \[ y’=\frac{\frac{x}{\sqrt{2(1+x^2)}}}{\frac{\sqrt{1-x^2}}{\sqrt{2}}} =\frac{x}{\sqrt{2(1+x^2)}}\cdot\frac{\sqrt{2}}{\sqrt{1-x^2}} =\frac{x}{\sqrt{(1+x^2)(1-x^2)}}. \] Since \[ \sqrt{(1+x^2)(1-x^2)}=\sqrt{1-x^4}, \] we obtain: \[ y’=\frac{x}{\sqrt{1-x^4}}. \]

Solution: Let \[ f(x)=\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right). \] Using the hint, substitute \[ x=\cos 2t. \] Then, we have: \[ \sqrt{1+x}=\sqrt{1+\cos 2t}=\sqrt{2\cos^2 t}=\sqrt{2}\cos t, \] \[ \sqrt{1-x}=\sqrt{1-\cos 2t}=\sqrt{2\sin^2 t}=\sqrt{2}\sin t, \] where we assume \( t \) is chosen so that \(\cos t\) and \(\sin t\) are nonnegative. Therefore, the expression inside the arctan becomes: \[ \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} =\frac{\sqrt{2}\cos t-\sqrt{2}\sin t}{\sqrt{2}\cos t+\sqrt{2}\sin t} =\frac{\cos t-\sin t}{\cos t+\sin t}. \] Dividing numerator and denominator by \(\cos t\) (assuming \(\cos t\ne0\)), we get: \[ \frac{1-\tan t}{1+\tan t}=\tan\left(\frac{\pi}{4}-t\right). \] Hence, \[ f(x)=\tan^{-1}\left(\tan\left(\frac{\pi}{4}-t\right)\right). \] Assuming \(\frac{\pi}{4}-t\) lies in the principal branch of \(\tan^{-1}\), we obtain: \[ f(x)=\frac{\pi}{4}-t. \] Now, since \( x=\cos 2t \), we have: \[ 2t=\cos^{-1}x \quad \Longrightarrow \quad t=\frac{1}{2}\cos^{-1}x. \] Substituting back, we find: \[ f(x)=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x. \] Differentiating with respect to \( x \): \[ f'(x)=-\frac{1}{2}\cdot\frac{d}{dx}\Big(\cos^{-1}x\Big) =-\frac{1}{2}\left(-\frac{1}{\sqrt{1-x^2}}\right) =\frac{1}{2\sqrt{1-x^2}}. \]

Solution: Let \[ f(x)=\sin^2\left(\cot^{-1}\sqrt{\frac{1+x}{1-x}}\right). \] To simplify, we use the hint by substituting \[ x=\cos 2t. \] First, consider the inner function: \[ u=\cot^{-1}\sqrt{\frac{1+x}{1-x}}. \] Rewriting \( u \) in terms of tangent, note that \[ \cot^{-1} z=\tan^{-1}\left(\frac{1}{z}\right). \] Therefore, \[ u=\tan^{-1}\left(\frac{1}{\sqrt{\frac{1+x}{1-x}}}\right) =\tan^{-1}\sqrt{\frac{1-x}{1+x}}. \] Now, substitute \( x=\cos 2t \): \[ \sqrt{1+x}=\sqrt{1+\cos 2t}=\sqrt{2\cos^2t}=\sqrt{2}\cos t, \] \[ \sqrt{1-x}=\sqrt{1-\cos 2t}=\sqrt{2\sin^2t}=\sqrt{2}\sin t. \] Hence, \[ \sqrt{\frac{1-x}{1+x}}=\frac{\sqrt{2}\sin t}{\sqrt{2}\cos t}=\tan t. \] Thus, \[ u=\tan^{-1}(\tan t)=t, \] where we assume \( t \) is in the principal range of \( \tan^{-1} \). Now, the original function becomes: \[ f(x)=\sin^2(u)=\sin^2t. \] But from the double-angle identity, note that when \( x=\cos 2t \), \[ \cos 2t = 1-2\sin^2t \quad \Longrightarrow \quad \sin^2t=\frac{1-x}{2}. \] Thus, we have: \[ f(x)=\frac{1-x}{2}. \] Differentiating with respect to \( x \): \[ f'(x)=\frac{d}{dx}\left(\frac{1-x}{2}\right) = -\frac{1}{2}. \]

Solution: Let \[ f(x)=\sin^{-1}\left(\frac{2\sec x}{1+\sec^2 x}\right). \] With the substitution \( \sec x=\tan t \), we obtain: \[ \frac{2\sec x}{1+\sec^2 x}=\frac{2\tan t}{1+\tan^2 t}=\sin 2t. \] Therefore, \[ f(x)=\sin^{-1}(\sin 2t)=2t, \] provided that \(2t\) lies within the principal range of \(\sin^{-1}\). Since \( t=\tan^{-1}(\sec x) \), it follows that: \[ f(x)=2\tan^{-1}(\sec x). \] Differentiating with respect to \( x \): \[ f'(x)=2\cdot\frac{d}{dx}\left(\tan^{-1}(\sec x)\right) =2\cdot\frac{1}{1+(\sec x)^2}\cdot\frac{d}{dx}(\sec x). \] As \[ \frac{d}{dx}(\sec x)=\sec x\tan x, \] we get: \[ f'(x)=\frac{2\sec x\tan x}{1+\sec^2 x}. \]

Solution: We start with the function \[ f(x)=\cos^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right). \] Notice that \[ \frac{\sin x+\cos x}{\sqrt{2}}=\sin\left(x+\frac{\pi}{4}\right), \] since \[ \sin\left(x+\frac{\pi}{4}\right)=\sin x\cos\frac{\pi}{4}+\cos x\sin\frac{\pi}{4}=\frac{\sin x+\cos x}{\sqrt{2}}. \] Hence, the function becomes \[ f(x)=\cos^{-1}\left(\sin\left(x+\frac{\pi}{4}\right)\right). \] Using the identity \[ \sin\theta=\cos\left(\frac{\pi}{2}-\theta\right), \] we write \[ \sin\left(x+\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{2}-x-\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}-x\right). \] Therefore, \[ f(x)=\cos^{-1}\left(\cos\left(\frac{\pi}{4}-x\right)\right). \] Provided that \(\frac{\pi}{4}-x\) lies within the principal value range of \(\cos^{-1}\), we have \[ f(x)=\frac{\pi}{4}-x. \] Differentiating with respect to \(x\) gives: \[ f'(x)=-1. \]

Solution: Let \[ f(x)=\cos^{-1}\left(\frac{3\cos x-4\sin x}{5}\right). \] Notice that the numerator can be written in the form \[ 3\cos x-4\sin x = 5\cos(x+\phi), \] where \(\phi\) is a constant angle satisfying \[ \cos\phi=\frac{3}{5} \quad \text{and} \quad \sin\phi=\frac{4}{5}. \] Hence, we have: \[ \frac{3\cos x-4\sin x}{5}=\cos(x+\phi). \] Therefore, \[ f(x)=\cos^{-1}\left(\cos(x+\phi)\right). \] Provided that \(x+\phi\) lies in the principal range \([0,\pi]\) of \(\cos^{-1}\), it follows that \[ f(x)=x+\phi. \] Differentiating with respect to \(x\): \[ f'(x)=1. \]

Solution: Let \[ f(x)=\sin^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right). \] To simplify the expression inside the inverse sine, substitute \[ x=\sin\theta, \] so that \[ \sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\cos\theta \quad (\text{assuming } \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). \] Then, the expression becomes: \[ \frac{x+\sqrt{1-x^2}}{\sqrt{2}}=\frac{\sin\theta+\cos\theta}{\sqrt{2}}. \] Recognize that: \[ \sin\theta+\cos\theta=\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right). \] Thus, \[ \frac{\sin\theta+\cos\theta}{\sqrt{2}}=\sin\left(\theta+\frac{\pi}{4}\right). \] Hence, we have: \[ f(x)=\sin^{-1}\left(\sin\left(\theta+\frac{\pi}{4}\right)\right). \] Provided that the angle \(\theta+\frac{\pi}{4}\) lies within the principal value range of \(\sin^{-1}\), we obtain: \[ f(x)=\theta+\frac{\pi}{4}. \] Since \( \theta=\sin^{-1}x \), it follows that: \[ f(x)=\sin^{-1}x+\frac{\pi}{4}. \] Differentiating with respect to \( x \): \[ f'(x)=\frac{d}{dx}\left(\sin^{-1}x+\frac{\pi}{4}\right) =\frac{1}{\sqrt{1-x^2}}. \]

Solution: Let \[ f(x)=\cos^{-1}\left(\frac{3x+4\sqrt{1-x^2}}{5}\right). \] Substitute \( x=\cos\theta \) so that \[ \sqrt{1-x^2}=\sqrt{1-\cos^2\theta}=\sin\theta. \] Then the expression inside the inverse cosine becomes: \[ \frac{3\cos\theta+4\sin\theta}{5}. \] Recognize that this can be written in the form: \[ 5\cos(\theta-\phi)=5\cos\theta\cos\phi+5\sin\theta\sin\phi, \] where we choose \(\phi\) such that \[ \cos\phi=\frac{3}{5} \quad \text{and} \quad \sin\phi=\frac{4}{5}. \] Thus, \[ 3\cos\theta+4\sin\theta=5\cos(\theta-\phi), \] and so \[ \frac{3x+4\sqrt{1-x^2}}{5}=\cos(\theta-\phi). \] Therefore, the function becomes: \[ f(x)=\cos^{-1}\left(\cos(\theta-\phi)\right). \] Assuming \(\theta-\phi\) lies in the principal range of \(\cos^{-1}\), we have: \[ f(x)=\theta-\phi. \] Since \( \theta=\cos^{-1}x \) and \(\phi=\cos^{-1}(3/5)\) (a constant), we can write: \[ f(x)=\cos^{-1}x-\cos^{-1}\left(\frac{3}{5}\right). \] Differentiating with respect to \( x \): \[ f'(x)=\frac{d}{dx}\Big(\cos^{-1}x\Big)= -\frac{1}{\sqrt{1-x^2}}, \] because the derivative of the constant term is zero.

Solution: Let \[ f(x)=\sin^{-1}\left(\frac{6x-4\sqrt{1-4x^2}}{5}\right). \] Observe that the expression inside the inverse sine can be rewritten using a sine difference identity. Define \[ u(x)=\frac{6x-4\sqrt{1-4x^2}}{5}. \] We aim to express \(u(x)\) in the form: \[ \sin\left(A-B\right)=\sin A\cos B-\cos A\sin B. \] Let \[ A=\sin^{-1}(2x) \quad \text{and} \quad B=\sin^{-1}\left(\frac{4}{5}\right). \] Then, \[ \sin A=2x,\quad \cos A=\sqrt{1-4x^2}, \] and \[ \sin B=\frac{4}{5},\quad \cos B=\frac{3}{5}. \] Using the sine difference formula: \[ \sin\left(A-B\right)=\sin A\cos B-\cos A\sin B =2x\cdot\frac{3}{5}-\sqrt{1-4x^2}\cdot\frac{4}{5} =\frac{6x-4\sqrt{1-4x^2}}{5}. \] Hence, we can write: \[ u(x)=\sin\left(\sin^{-1}(2x)-\sin^{-1}\left(\frac{4}{5}\right)\right). \] Taking the inverse sine on both sides (and noting that the inverse sine of a sine returns the angle within the principal branch), we have: \[ f(x)=\sin^{-1}\big(u(x)\big) =\sin^{-1}(2x)-\sin^{-1}\left(\frac{4}{5}\right). \] Since \(\sin^{-1}\left(\frac{4}{5}\right)\) is a constant, when differentiating \(f(x)\) with respect to \(x\) we obtain: \[ f'(x)=\frac{d}{dx}\Big(\sin^{-1}(2x)\Big) =\frac{2}{\sqrt{1-(2x)^2}} =\frac{2}{\sqrt{1-4x^2}}. \]

Solution: Let \[ f(x)=\tan^{-1}\left(x^{1/3}\right). \] Set \[ u(x)=x^{1/3}. \] Then, by the chain rule: \[ f'(x)=\frac{u'(x)}{1+u(x)^2}. \] First, differentiate \( u(x)=x^{1/3} \): \[ u'(x)=\frac{1}{3}x^{-2/3}. \] Also, note that: \[ u(x)^2=x^{2/3}. \] Therefore, \[ f'(x)=\frac{\frac{1}{3}x^{-2/3}}{1+x^{2/3}} =\frac{1}{3x^{2/3}(1+x^{2/3})}. \]

Solution: First, recognize that the expression can be simplified using a well-known trigonometric identity. Recall that: \[ \sin\left(2\sin^{-1} x\right)=2x\sqrt{1-x^2}. \] Let \[ f(x)=2x\sqrt{1-x^2}. \] To differentiate \(f(x)\), we use the product rule. Define: \[ u(x)=2x \quad \text{and} \quad v(x)=\sqrt{1-x^2}. \] Then, \[ f'(x)=u'(x)v(x)+u(x)v'(x). \] First, compute: \[ u'(x)=2. \] Next, write \(v(x)=(1-x^2)^{1/2}\) and differentiate using the chain rule: \[ v'(x)=\frac{1}{2}(1-x^2)^{-1/2}\cdot(-2x)=-\frac{x}{\sqrt{1-x^2}}. \] Now, substitute back: \[ f'(x)=2\sqrt{1-x^2}+2x\left(-\frac{x}{\sqrt{1-x^2}}\right) =2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}. \] Combine the terms over a common denominator: \[ f'(x)=\frac{2(1-x^2)-2x^2}{\sqrt{1-x^2}} =\frac{2(1-2x^2)}{\sqrt{1-x^2}}. \]

Solution: First, simplify the expression inside the inverse cotangent. Define: \[ A = \sqrt{1+\sin x} \quad \text{and} \quad B = \sqrt{1-\sin x}. \] Using the half-angle identities, we have: \[ \sqrt{1+\sin x} = \sin\frac{x}{2}+\cos\frac{x}{2} \quad \text{and} \quad \sqrt{1-\sin x} = \cos\frac{x}{2}-\sin\frac{x}{2}. \] Therefore, \[ A+B = (\sin\frac{x}{2}+\cos\frac{x}{2})+(\cos\frac{x}{2}-\sin\frac{x}{2}) = 2\cos\frac{x}{2}, \] and \[ A-B = (\sin\frac{x}{2}+\cos\frac{x}{2})-(\cos\frac{x}{2}-\sin\frac{x}{2}) = 2\sin\frac{x}{2}. \] Hence, the given expression simplifies to: \[ \frac{A+B}{A-B} = \frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}. \] This implies: \[ y = \cot^{-1}\left(\cot\frac{x}{2}\right). \] Assuming \(\frac{x}{2}\) lies within the principal range of \(\cot^{-1}\), we obtain: \[ y = \frac{x}{2}. \] Differentiating with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{2}. \]