Ex 5.12 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: We differentiate both \(x\) and \(y\) with respect to \(t\): \[ \frac{dx}{dt} = \frac{d}{dt}(a t^2) = 2a t, \] \[ \frac{dy}{dt} = \frac{d}{dt}(2a t) = 2a. \] Using the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2a t} = \frac{1}{t}. \]

Solution: Differentiate both functions with respect to \(t\): \[ \frac{dx}{dt} = \frac{d}{dt}(2a t^2) = 4a t, \] \[ \frac{dy}{dt} = \frac{d}{dt}(a t^4) = 4a t^3. \] Then, using the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4a t^3}{4a t} = t^2. \]

Solution: Differentiate with respect to \(\theta\): \[ \frac{dx}{d\theta} = -a \sin \theta,\quad \frac{dy}{d\theta} = -b \sin \theta. \] Then, by the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-b \sin \theta}{-a \sin \theta} = \frac{b}{a}. \]

Solution: Differentiate both functions with respect to \(\theta\): \[ \frac{dx}{d\theta} = -a \sin \theta,\quad \frac{dy}{d\theta} = a \cos \theta. \] Using the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta. \]

Solution: Differentiate both functions with respect to \(\theta\): \[ \frac{dx}{d\theta} = a \sec \theta \tan \theta,\quad \frac{dy}{d\theta} = b \sec^2 \theta. \] Then, using the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{b \sec^2 \theta}{a \sec \theta \tan \theta} = \frac{b \sec \theta}{a \tan \theta}. \] Express \(\sec \theta\) and \(\tan \theta\) in terms of sine: \[ \sec \theta = \frac{1}{\cos \theta},\quad \tan \theta = \frac{\sin \theta}{\cos \theta}. \] Substituting these: \[ \frac{dy}{dx} = \frac{b \cdot \frac{1}{\cos \theta}}{a \cdot \frac{\sin \theta}{\cos \theta}} = \frac{b}{a \sin \theta} = \frac{b}{a}\,\mathrm{cosec}\,\theta. \]

Solution: Differentiate both functions with respect to \(t\): \[ \frac{dx}{dt} = \cos t,\quad \frac{dy}{dt} = -2\sin 2t. \] Since \(\sin 2t = 2\sin t\cos t\), we have: \[ \frac{dy}{dt} = -2(2\sin t\cos t) = -4\sin t\cos t. \] Therefore, applying the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-4\sin t\cos t}{\cos t} = -4\sin t. \]

Solution: Differentiate both functions with respect to \(t\): \[ \frac{dx}{dt} = 4, \] \[ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{4}{t}\right) = -\frac{4}{t^2}. \] Then, using the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{4}{t^2}}{4} = -\frac{1}{t^2}. \]

Solution: First, differentiate \(x\) and \(y\) with respect to \(t\): \[ \frac{dx}{dt} = 1 – \frac{1}{t^2}, \] \[ \frac{dy}{dt} = 1 + \frac{1}{t^2}. \] Then, applying the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1 + \frac{1}{t^2}}{1 – \frac{1}{t^2}} = \frac{t^2+1}{t^2-1}. \]

Solution: First, differentiate with respect to \(t\): \[ \frac{dx}{dt} = a(1-\cos t), \] \[ \frac{dy}{dt} = -a\sin t. \] Then, by the chain rule: \[ \frac{dy}{dx} = \frac{-a\sin t}{a(1-\cos t)} = -\frac{\sin t}{1-\cos t}. \] Using the half-angle identities: \[ \sin t = 2\sin\frac{t}{2}\cos\frac{t}{2},\quad 1-\cos t = 2\sin^2\frac{t}{2}, \] we have: \[ \frac{dy}{dx} = -\frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}} = -\cot\frac{t}{2}. \]

Solution: Differentiate \(x\) and \(y\) with respect to \(t\): \[ \frac{dx}{dt} = \frac{d}{dt}\left(a \sin^3 t\right) = 3a \sin^2 t \cos t, \] \[ \frac{dy}{dt} = \frac{d}{dt}\left(a \cos^3 t\right) = -3a \cos^2 t \sin t. \] Then, by the chain rule: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-3a \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\frac{\cos t}{\sin t} = -\cot t. \]

Solution: First, we differentiate the given functions with respect to \(\theta\): \[ \frac{dx}{d\theta} = a\left(1+\cos \theta\right) \] \[ \frac{dy}{d\theta} = a\sin \theta \] Now, the derivative \(\frac{dy}{dx}\) is given by the ratio: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\sin \theta}{a(1+\cos \theta)} = \frac{\sin \theta}{1+\cos \theta}. \] Using the half-angle identities: \[ \sin \theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \quad \text{and} \quad 1+\cos \theta = 2\cos^2\frac{\theta}{2}, \] we simplify: \[ \frac{dy}{dx} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}. \]

Solution: First, differentiate the given functions with respect to \(\theta\): \[ \frac{dx}{d\theta} = a\sin \theta, \] \[ \frac{dy}{d\theta} = a\left(1+\cos \theta\right). \] Thus, the derivative \(\frac{dy}{dx}\) is: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a(1+\cos \theta)}{a\sin \theta} = \frac{1+\cos \theta}{\sin \theta}. \] Using the half-angle identities: \[ 1+\cos \theta = 2\cos^2\frac{\theta}{2} \quad \text{and} \quad \sin \theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, \] we simplify: \[ \frac{dy}{dx} = \frac{2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} = \cot\frac{\theta}{2}. \]

Solution: The given functions are: \[ x=e^{t}(\sin t+\cos t) \quad \text{and} \quad y=e^{t}(\sin t-\cos t). \] Differentiate \( x \) with respect to \( t \) using the product rule: \[ \frac{dx}{dt} = e^{t}(\sin t+\cos t) + e^{t}(\cos t-\sin t) \] Simplifying: \[ \frac{dx}{dt} = e^{t}\Big[(\sin t+\cos t)+(\cos t-\sin t)\Big] = e^{t}(2\cos t) = 2e^{t}\cos t. \] Next, differentiate \( y \) with respect to \( t \): \[ \frac{dy}{dt} = e^{t}(\sin t-\cos t) + e^{t}(\cos t+\sin t) \] Simplifying: \[ \frac{dy}{dt} = e^{t}\Big[(\sin t-\cos t)+(\cos t+\sin t)\Big] = e^{t}(2\sin t) = 2e^{t}\sin t. \] Now, the derivative \(\frac{dy}{dx}\) is given by: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2e^{t}\sin t}{2e^{t}\cos t} = \frac{\sin t}{\cos t} = \tan t. \]

Solution: The given functions are: \[ x=\cos \theta-\cos 2\theta, \quad y=\sin \theta-\sin 2\theta. \] Differentiate \( x \) with respect to \( \theta \): \[ \frac{dx}{d\theta} = -\sin \theta + 2\sin 2\theta. \] Differentiate \( y \) with respect to \( \theta \): \[ \frac{dy}{d\theta} = \cos \theta – 2\cos 2\theta. \] Therefore, the derivative \(\frac{dy}{dx}\) is: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\cos \theta – 2\cos 2\theta}{2\sin 2\theta – \sin \theta}. \]

Solution: The given functions are: \[ x=a\left(\cos \theta+\cos 2\theta\right) \quad \text{and} \quad y=b\left(\sin \theta+\sin 2\theta\right). \] Differentiating \( x \) with respect to \(\theta\): \[ \frac{dx}{d\theta}=a\left(-\sin \theta-2\sin 2\theta\right). \] Differentiating \( y \) with respect to \(\theta\): \[ \frac{dy}{d\theta}=b\left(\cos \theta+2\cos 2\theta\right). \] Therefore, the derivative \(\frac{dy}{dx}\) is: \[ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{b\left(\cos \theta+2\cos 2\theta\right)}{a\left(-\sin \theta-2\sin 2\theta\right)}=-\frac{b}{a}\cdot\frac{\cos \theta+2\cos 2\theta}{\sin \theta+2\sin 2\theta}. \]

Solution: We have \[ x=\frac{1+\log t}{t^{2}} \quad \text{and} \quad y=\frac{3+2\log t}{t}. \] Express \( x \) and \( y \) as: \[ x=(1+\log t)t^{-2}, \quad y=(3+2\log t)t^{-1}. \] Differentiate \( x \) with respect to \( t \) using the product rule: \[ \frac{dx}{dt} = \frac{d}{dt}(1+\log t)\cdot t^{-2} + (1+\log t)\cdot\frac{d}{dt}(t^{-2}). \] Here, \[ \frac{d}{dt}(1+\log t)=\frac{1}{t} \quad \text{and} \quad \frac{d}{dt}(t^{-2})=-2t^{-3}. \] Thus, \[ \frac{dx}{dt} = \frac{1}{t}\cdot t^{-2} + (1+\log t)(-2t^{-3}) = t^{-3}-2(1+\log t)t^{-3}. \] Simplify: \[ \frac{dx}{dt} = \frac{1-2-2\log t}{t^{3}}=\frac{-1-2\log t}{t^{3}}. \] Next, differentiate \( y \) with respect to \( t \): \[ \frac{dy}{dt} = \frac{d}{dt}(3+2\log t)\cdot t^{-1} + (3+2\log t)\cdot\frac{d}{dt}(t^{-1}). \] Here, \[ \frac{d}{dt}(3+2\log t)=\frac{2}{t} \quad \text{and} \quad \frac{d}{dt}(t^{-1})=-t^{-2}. \] Thus, \[ \frac{dy}{dt} = \frac{2}{t}\cdot t^{-1} – (3+2\log t)t^{-2} = \frac{2}{t^{2}}-\frac{3+2\log t}{t^{2}}. \] Simplify: \[ \frac{dy}{dt} = \frac{2-3-2\log t}{t^{2}}=\frac{-1-2\log t}{t^{2}}. \] Now, the derivative \(\frac{dy}{dx}\) is: \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-1-2\log t}{t^{2}}}{\frac{-1-2\log t}{t^{3}}}=\frac{-1-2\log t}{t^{2}} \cdot \frac{t^{3}}{-1-2\log t} = t. \]

Solution: We are given: \[ x=3\cos \theta-2\cos^{3}\theta, \quad y=3\sin \theta-2\sin^{3}\theta. \] Differentiating \( x \) with respect to \(\theta\): \[ \frac{dx}{d\theta}=\frac{d}{d\theta}(3\cos \theta)-2\frac{d}{d\theta}(\cos^{3}\theta). \] Compute each derivative: \[ \frac{d}{d\theta}(3\cos \theta)=-3\sin \theta, \] and using the chain rule, \[ \frac{d}{d\theta}(\cos^{3}\theta)=3\cos^{2}\theta\cdot (-\sin \theta)=-3\cos^{2}\theta\sin \theta. \] Hence, \[ \frac{dx}{d\theta}=-3\sin \theta-2\left(-3\cos^{2}\theta\sin \theta\right)=-3\sin \theta+6\cos^{2}\theta\sin \theta. \] Factorizing: \[ \frac{dx}{d\theta}=\sin \theta\left(6\cos^{2}\theta-3\right). \] Next, differentiate \( y \) with respect to \(\theta\): \[ \frac{dy}{d\theta}=\frac{d}{d\theta}(3\sin \theta)-2\frac{d}{d\theta}(\sin^{3}\theta). \] Compute each derivative: \[ \frac{d}{d\theta}(3\sin \theta)=3\cos \theta, \] and \[ \frac{d}{d\theta}(\sin^{3}\theta)=3\sin^{2}\theta\cos \theta. \] Thus, \[ \frac{dy}{d\theta}=3\cos \theta-2\cdot 3\sin^{2}\theta\cos \theta=3\cos \theta-6\sin^{2}\theta\cos \theta. \] Factorizing: \[ \frac{dy}{d\theta}=\cos \theta\left(3-6\sin^{2}\theta\right). \] Now, the derivative \(\frac{dy}{dx}\) is given by: \[ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\cos \theta\left(3-6\sin^{2}\theta\right)}{\sin \theta\left(6\cos^{2}\theta-3\right)}. \] Notice that: \[ 6\cos^{2}\theta-3=6(1-\sin^{2}\theta)-3=6-6\sin^{2}\theta-3=3-6\sin^{2}\theta. \] Therefore, we have: \[ \frac{dy}{dx}=\frac{\cos \theta\left(3-6\sin^{2}\theta\right)}{\sin \theta\left(3-6\sin^{2}\theta\right)}=\frac{\cos \theta}{\sin \theta}=\cot \theta. \]

Solution: We are given: \[ x=2\cos t-\cos 2t, \quad y=2\sin t-\sin 2t. \] Differentiate \( x \) with respect to \( t \): \[ \frac{dx}{dt}=-2\sin t+2\sin 2t. \] Differentiate \( y \) with respect to \( t \): \[ \frac{dy}{dt}=2\cos t-2\cos 2t. \] Thus, the derivative is: \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2\cos t-2\cos 2t}{-2\sin t+2\sin 2t}=\frac{\cos t-\cos 2t}{-\sin t+\sin 2t}. \] Substituting \( t=\frac{\pi}{4} \): \[ \cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \cos\frac{\pi}{2}=0, \] \[ \sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \sin\frac{\pi}{2}=1. \] Therefore, \[ \frac{dy}{dx}=\frac{\frac{\sqrt{2}}{2}-0}{-\frac{\sqrt{2}}{2}+1}=\frac{\frac{\sqrt{2}}{2}}{\frac{2-\sqrt{2}}{2}}=\frac{\sqrt{2}}{2-\sqrt{2}}. \] Multiplying numerator and denominator by \(2+\sqrt{2}\): \[ \frac{dy}{dx}=\frac{\sqrt{2}(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}=\frac{\sqrt{2}(2+\sqrt{2})}{4-2}=\frac{\sqrt{2}(2+\sqrt{2})}{2}. \] Simplify further: \[ \frac{dy}{dx}=\frac{2\sqrt{2}+2}{2}=\sqrt{2}+1. \]

Solution: The given functions are: \[ x=3\sin t-\sin 3t \quad \text{and} \quad y=3\cos t-\cos 3t. \] Differentiating \( x \) with respect to \( t \): \[ \frac{dx}{dt} = 3\cos t – 3\cos 3t. \] Differentiating \( y \) with respect to \( t \): \[ \frac{dy}{dt} = -3\sin t + 3\sin 3t. \] Hence, the derivative \(\frac{dy}{dx}\) is: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-3\sin t+3\sin 3t}{3\cos t-3\cos 3t} = \frac{-\sin t+\sin 3t}{\cos t-\cos 3t}. \] Now, substituting \( t=\frac{\pi}{3} \): \[ \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}, \quad \cos\frac{\pi}{3} = \frac{1}{2}, \] \[ \sin\left(3\cdot\frac{\pi}{3}\right)=\sin\pi = 0, \quad \cos\left(3\cdot\frac{\pi}{3}\right)=\cos\pi = -1. \] Therefore, \[ \frac{dy}{dx} = \frac{-\frac{\sqrt{3}}{2}+0}{\frac{1}{2}-(-1)} = \frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}} = -\frac{\sqrt{3}}{3}. \] Noting that \(-\frac{\sqrt{3}}{3}\) is equivalent to \(-\frac{1}{\sqrt{3}}\), we have our final result.

Solution: We are given: \[ x=\frac{2b t}{1+t^{2}} \quad \text{and} \quad y=\frac{a(1-t^{2})}{1+t^{2}}. \] \(\underline{\text{Differentiate } x \text{ w.r.t. } t:}\) Using the quotient rule, for \[ x=\frac{2bt}{1+t^{2}}, \] we have: \[ \frac{dx}{dt}=\frac{(1+t^{2})\cdot (2b)-2bt\cdot (2t)}{(1+t^{2})^{2}}. \] Simplify the numerator: \[ (1+t^{2})\cdot (2b)-2bt\cdot (2t)=2b(1+t^{2})-4bt^{2}=2b\left(1+t^{2}-2t^{2}\right)=2b(1-t^{2}). \] Therefore, \[ \frac{dx}{dt}=\frac{2b(1-t^{2})}{(1+t^{2})^{2}}. \] \(\underline{\text{Differentiate } y \text{ w.r.t. } t:}\) For \[ y=\frac{a(1-t^{2})}{1+t^{2}}, \] applying the quotient rule: \[ \frac{dy}{dt}=\frac{(1+t^{2})\cdot \frac{d}{dt}\bigl(a(1-t^{2})\bigr)-a(1-t^{2})\cdot \frac{d}{dt}(1+t^{2})}{(1+t^{2})^{2}}. \] Compute the derivatives: \[ \frac{d}{dt}\bigl(a(1-t^{2})\bigr)=-2at, \quad \frac{d}{dt}(1+t^{2})=2t. \] Substitute these into the expression: \[ \frac{dy}{dt}=\frac{(1+t^{2})(-2at)-a(1-t^{2})(2t)}{(1+t^{2})^{2}}. \] Simplify the numerator: \[ -2at(1+t^{2})-2at(1-t^{2})=-2at\bigl[(1+t^{2})+(1-t^{2})\bigr]=-2at(2)= -4at. \] Thus, \[ \frac{dy}{dt}=\frac{-4at}{(1+t^{2})^{2}}. \] \(\underline{\text{Find } \frac{dy}{dx}:}\) \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\displaystyle \frac{-4at}{(1+t^{2})^{2}}}{\displaystyle \frac{2b(1-t^{2})}{(1+t^{2})^{2}}}=\frac{-4at}{2b(1-t^{2})}=\frac{-2at}{b(1-t^{2})}. \] \(\underline{\text{Substitute } t=2:}\) \[ \frac{dy}{dx}\Biggr|_{t=2}=\frac{-2a(2)}{b\bigl(1-2^{2}\bigr)}=\frac{-4a}{b(1-4)}=\frac{-4a}{b(-3)}=\frac{4a}{3b}. \]

Solution: Let \( u = x^2 \). Then the given function can be written as \[ f(u) = \frac{u}{1-u}. \] Differentiating \( f(u) \) with respect to \( u \) using the quotient rule: \[ \frac{d f}{du} = \frac{(1-u)\cdot 1 – u\cdot (-1)}{(1-u)^2} = \frac{1-u+u}{(1-u)^2} = \frac{1}{(1-u)^2}. \] Replacing \( u \) back with \( x^2 \), we get: \[ \frac{d}{d(x^2)}\left(\frac{x^{2}}{1-x^{2}}\right) = \frac{1}{\left(1-x^{2}\right)^2}. \]

Solution: Let \( u = x^3 \). To find \(\frac{d}{du}(\sin x^2)\), we use the chain rule: \[ \frac{d}{du}(\sin x^2)= \frac{d}{dx}(\sin x^2) \cdot \frac{dx}{du}. \] First, differentiate \(\sin x^2\) with respect to \(x\): \[ \frac{d}{dx}(\sin x^2)= \cos x^2 \cdot 2x. \] Next, differentiate \(x^3\) with respect to \(x\): \[ \frac{du}{dx} = 3x^2 \quad \Longrightarrow \quad \frac{dx}{du} = \frac{1}{3x^2}. \] Hence, \[ \frac{d}{du}(\sin x^2)= \left(2x \cos x^2\right) \cdot \frac{1}{3x^2} = \frac{2\cos x^2}{3x}. \]

Solution: Let \( f(x)=\cot^{3}(2x+1) \) and \( u=x^2+1 \). Then, by the chain rule, \[ \frac{d f}{du}=\frac{\frac{d f}{dx}}{\frac{d u}{dx}}. \] First, differentiate \( u=x^2+1 \): \[ \frac{d u}{dx}=2x. \] Next, differentiate \( f(x) \): \[ f(x)=\left[\cot(2x+1)\right]^3. \] Using the chain rule: \[ \frac{d f}{dx}=3\left[\cot(2x+1)\right]^2\cdot \frac{d}{dx}\left[\cot(2x+1)\right]. \] We have: \[ \frac{d}{dx}\left[\cot(2x+1)\right] = -\mathrm{cosec}^{2}(2x+1)\cdot (2), \] so: \[ \frac{d f}{dx} = 3\left[\cot(2x+1)\right]^2\cdot\left(-2\,\mathrm{cosec}^{2}(2x+1)\right) = -6\left[\cot(2x+1)\right]^2\,\mathrm{cosec}^{2}(2x+1). \] Therefore, \[ \frac{d f}{du}=\frac{-6\left[\cot(2x+1)\right]^2\,\mathrm{cosec}^{2}(2x+1)}{2x} = -\frac{3\left[\cot(2x+1)\right]^2\,\mathrm{cosec}^{2}(2x+1)}{x}. \]

Solution: Let \[ u = \sqrt{\cos x} = (\cos x)^{\frac{1}{2}}, \] and the function is \[ f(x)=\log (\sin x). \] We use the chain rule: \[ \frac{d f}{d u} = \frac{\frac{d f}{d x}}{\frac{d u}{d x}}. \] First, differentiate \(f(x)\) with respect to \(x\): \[ \frac{d}{dx}\left(\log (\sin x)\right) = \frac{\cos x}{\sin x} = \cot x. \] Next, differentiate \(u\) with respect to \(x\): \[ \frac{d u}{dx} = \frac{1}{2}(\cos x)^{-\frac{1}{2}}(-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}. \] Thus, \[ \frac{d f}{d u} = \frac{\cot x}{-\frac{\sin x}{2\sqrt{\cos x}}} = -2\sqrt{\cos x}\cdot \frac{\cot x}{\sin x}. \] Since \(\cot x = \frac{\cos x}{\sin x}\), we have: \[ \frac{d f}{d u} = -2\sqrt{\cos x}\cdot \frac{\cos x}{\sin^2 x} = -2\sqrt{\cos x}\, \cot x\, \mathrm{cosec}\, x. \]

Solution: Let \[ y=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right) \quad \text{and} \quad u=\tan^{-1}x. \] We use the chain rule: \[ \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}}. \] \(\underline{\text{Step 1: Compute } \frac{du}{dx}}\) Since \[ u=\tan^{-1}x,\quad \frac{du}{dx}=\frac{1}{1+x^{2}}. \] \(\underline{\text{Step 2: Compute } \frac{dy}{dx}}\) Write \[ y=\sin^{-1}(z) \quad \text{where} \quad z=\frac{2x}{1+x^{2}}. \] Then, \[ \frac{dy}{dz}=\frac{1}{\sqrt{1-z^{2}}}. \] Differentiate \(z\) with respect to \(x\) using the quotient rule: \[ \frac{dz}{dx}=\frac{(1+x^{2})\cdot 2-2x\cdot 2x}{(1+x^{2})^{2}} =\frac{2(1+x^{2})-4x^{2}}{(1+x^{2})^{2}} =\frac{2(1-x^{2})}{(1+x^{2})^{2}}. \] Next, compute \(1-z^{2}\): \[ z^{2}=\frac{4x^{2}}{(1+x^{2})^{2}},\quad 1-z^{2}=\frac{(1+x^{2})^{2}-4x^{2}}{(1+x^{2})^{2}} =\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}. \] Therefore, \[ \sqrt{1-z^{2}}=\frac{|1-x^{2}|}{1+x^{2}}. \] Assuming \(|x|<1\) so that \(1-x^{2}>0\), we have: \[ \sqrt{1-z^{2}}=\frac{1-x^{2}}{1+x^{2}}. \] Then, \[ \frac{dy}{dz}=\frac{1+x^{2}}{1-x^{2}}. \] Hence, \[ \frac{dy}{dx}=\frac{dy}{dz}\cdot \frac{dz}{dx} =\frac{1+x^{2}}{1-x^{2}} \cdot \frac{2(1-x^{2})}{(1+x^{2})^{2}} =\frac{2}{1+x^{2}}. \] \(\underline{\text{Step 3: Compute } \frac{dy}{du}}\) Using \[ \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}} =\frac{\frac{2}{1+x^{2}}}{\frac{1}{1+x^{2}}} =2. \]

Solution: Let \[ y=\tan^{-1}\left(\frac{2x}{1-x^{2}}\right) \quad \text{and} \quad u=\tan^{-1}x. \] We use the chain rule: \[ \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}}. \] \(\underline{\text{Step 1: Compute } \frac{du}{dx}}\) Since \[ u=\tan^{-1}x,\quad \frac{du}{dx}=\frac{1}{1+x^{2}}. \] \(\underline{\text{Step 2: Compute } \frac{dy}{dx}}\) Write \[ y=\tan^{-1}(z) \quad \text{where} \quad z=\frac{2x}{1-x^{2}}. \] Then, \[ \frac{dy}{dz}=\frac{1}{1+z^{2}}. \] Differentiating \(z\) with respect to \(x\) using the quotient rule: \[ \frac{dz}{dx}=\frac{(1-x^{2})\cdot 2 – 2x\cdot(-2x)}{(1-x^{2})^{2}} =\frac{2(1-x^{2})+4x^{2}}{(1-x^{2})^{2}} =\frac{2(1+x^{2})}{(1-x^{2})^{2}}. \] Next, compute \(1+z^{2}\): \[ z^{2}=\frac{4x^{2}}{(1-x^{2})^{2}},\quad 1+z^{2}=\frac{(1-x^{2})^{2}+4x^{2}}{(1-x^{2})^{2}} =\frac{1+2x^{2}+x^{4}}{(1-x^{2})^{2}} =\frac{(1+x^{2})^{2}}{(1-x^{2})^{2}}. \] Thus, \[ \frac{dy}{dz}=\frac{1}{1+z^{2}}=\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}}. \] Hence, \[ \frac{dy}{dx}=\frac{dy}{dz}\cdot \frac{dz}{dx} =\frac{(1-x^{2})^{2}}{(1+x^{2})^{2}} \cdot \frac{2(1+x^{2})}{(1-x^{2})^{2}} =\frac{2}{1+x^{2}}. \] \(\underline{\text{Step 3: Compute } \frac{dy}{du}}\) Using \[ \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}} =\frac{\frac{2}{1+x^{2}}}{\frac{1}{1+x^{2}}} =2. \]

Solution: Let \( x = \sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right) \) and \( y = \cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right) \). Consider, \[ \sin x = \frac{t}{\sqrt{1+t^{2}}} \] Then, \[ \cos x = \sqrt{1 – \sin^2 x} = \sqrt{1 – \left( \frac{t}{\sqrt{1+t^{2}}} \right)^2} = \sqrt{ \frac{1+t^{2} – t^{2}}{1+t^{2}} } = \sqrt{ \frac{1}{1+t^{2}} } = \frac{1}{\sqrt{1+t^{2}}} \] Therefore, \[ \cos^{-1}\left( \frac{1}{\sqrt{1+t^{2}}} \right) = x = \sin^{-1}\left( \frac{t}{\sqrt{1+t^{2}}} \right) \] Hence, \[ y = x \Rightarrow \frac{dy}{dx} = 1 \]

Solution: Let \[ y=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right) \quad \text{and} \quad u=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right). \] We differentiate \(y\) and \(u\) with respect to \(x\) and then form the ratio: \[ \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}}. \] A key observation is obtained by substituting \[ x=\tan \theta. \] Then, using the double-angle formulas: \[ \sin y=\frac{2\tan \theta}{1+\tan^{2}\theta}=\sin(2\theta) \quad \Longrightarrow \quad y=2\theta, \] and \[ \cos u=\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\cos(2\theta) \quad \Longrightarrow \quad u=2\theta. \] Thus, we have \[ y=u. \] Differentiating \(y\) with respect to \(u\) gives: \[ \frac{dy}{du}=1. \]

Solution: Let \[ y=\tan^{-1}\frac{3x-x^{3}}{1-3x^{2}} \quad \text{and} \quad u=\tan^{-1}\frac{2x}{1-x^{2}}. \] Substitute \[ x=\tan\theta. \] Then, using the tangent multiple-angle formulas, we have: \[ \frac{3x-x^{3}}{1-3x^{2}}=\frac{3\tan\theta-\tan^{3}\theta}{1-3\tan^{2}\theta}=\tan(3\theta), \] and \[ \frac{2x}{1-x^{2}}=\frac{2\tan\theta}{1-\tan^{2}\theta}=\tan(2\theta). \] Therefore, \[ y=\tan^{-1}\big(\tan 3\theta\big)=3\theta \quad \text{and} \quad u=\tan^{-1}\big(\tan 2\theta\big)=2\theta. \] Expressing \(y\) in terms of \(u\): \[ y=\frac{3}{2}u. \] Differentiating with respect to \(u\): \[ \frac{dy}{du}=\frac{3}{2}. \]

Solution: Let \[ y=\tan^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right) \quad \text{and} \quad u=\tan^{-1}x. \] Then by the chain rule, \[ \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}}. \] First, we have \[ \frac{du}{dx}=\frac{1}{1+x^{2}}. \] Next, write \[ y=\tan^{-1}(z),\quad \text{where} \quad z=\frac{\sqrt{1+x^{2}}-1}{x}. \] Then, \[ \frac{dy}{dz}=\frac{1}{1+z^{2}}. \] To find \(\frac{dz}{dx}\), note that \[ z=\frac{\sqrt{1+x^{2}}-1}{x}. \] Differentiating the numerator: \[ \frac{d}{dx}\left(\sqrt{1+x^{2}}-1\right)=\frac{x}{\sqrt{1+x^{2}}}. \] Using the quotient rule, \[ \frac{dz}{dx}=\frac{x\left(\frac{x}{\sqrt{1+x^{2}}}\right)-\left(\sqrt{1+x^{2}}-1\right)}{x^{2}} =\frac{\frac{x^{2}}{\sqrt{1+x^{2}}}-\left(\sqrt{1+x^{2}}-1\right)}{x^{2}}. \] Simplify the numerator by writing it as: \[ \frac{x^{2}}{\sqrt{1+x^{2}}}-\sqrt{1+x^{2}}+1 =\frac{x^{2}-(1+x^{2})+ \sqrt{1+x^{2}}}{\sqrt{1+x^{2}}} =\frac{-1+\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}}. \] Thus, \[ \frac{dz}{dx}=\frac{1-\frac{1}{\sqrt{1+x^{2}}}}{x^{2}}. \] Therefore, \[ \frac{dy}{dx}=\frac{dy}{dz}\cdot \frac{dz}{dx} =\frac{1}{1+z^{2}}\cdot \frac{1-\frac{1}{\sqrt{1+x^{2}}}}{x^{2}}. \] Now, compute \(z^{2}\): \[ z^{2}=\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)^{2} =\frac{(\sqrt{1+x^{2}}-1)^{2}}{x^{2}}. \] Notice that \[ (\sqrt{1+x^{2}}-1)^{2}=1+x^{2}-2\sqrt{1+x^{2}}+1 =x^{2}+2-2\sqrt{1+x^{2}}. \] Hence, \[ 1+z^{2}=\frac{2x^{2}+2-2\sqrt{1+x^{2}}}{x^{2}} =\frac{2\left(x^{2}+1-\sqrt{1+x^{2}}\right)}{x^{2}}. \] Thus, \[ \frac{1}{1+z^{2}}=\frac{x^{2}}{2\left(x^{2}+1-\sqrt{1+x^{2}}\right)}. \] Substituting back, we obtain: \[ \frac{dy}{dx}=\frac{x^{2}}{2\left(x^{2}+1-\sqrt{1+x^{2}}\right)}\cdot \frac{1-\frac{1}{\sqrt{1+x^{2}}}}{x^{2}} =\frac{1-\frac{1}{\sqrt{1+x^{2}}}}{2\left(x^{2}+1-\sqrt{1+x^{2}}\right)}. \] Now, using the chain rule, \[ \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}} =\frac{\frac{1-\frac{1}{\sqrt{1+x^{2}}}}{2\left(x^{2}+1-\sqrt{1+x^{2}}\right)}}{\frac{1}{1+x^{2}}} =\frac{1-\frac{1}{\sqrt{1+x^{2}}}}{2\left(x^{2}+1-\sqrt{1+x^{2}}\right)}\cdot (1+x^{2}). \] Notice that \(1+x^{2}=(\sqrt{1+x^{2}})^{2}\). Hence, \[ \frac{dy}{du}=\frac{(\sqrt{1+x^{2}})^{2}\left(1-\frac{1}{\sqrt{1+x^{2}}}\right)}{2\left((\sqrt{1+x^{2}})^{2}-\sqrt{1+x^{2}}\right)} =\frac{\sqrt{1+x^{2}}\left(\sqrt{1+x^{2}}-1\right)}{2\sqrt{1+x^{2}}\left(\sqrt{1+x^{2}}-1\right)} =\frac{1}{2}. \]

Solution: Let \[ u(x)= \tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right) \quad\text{and}\quad v(x)= \cos^{-1}\left(2x\sqrt{1-x^{2}}\right). \] Using the substitution \( x=\cos \theta \) (so that \(\sqrt{1-x^{2}}=\sin \theta\)), we obtain: \[ u(x)= \tan^{-1}\left(\frac{\sin \theta}{\cos \theta}\right)= \theta = \cos^{-1}(x). \] Hence, \[ u'(x)= \frac{d}{dx}\cos^{-1}(x) = -\frac{1}{\sqrt{1-x^{2}}}. \] Also, \[ v(x)= \cos^{-1}\left(2\cos \theta \sin \theta\right) = \cos^{-1}\left(\sin 2\theta\right). \] Since \(\sin 2\theta=\cos\left(\frac{\pi}{2}-2\theta\right)\) and assuming the angle \(\frac{\pi}{2}-2\theta\) lies in \([0,\pi]\), we have: \[ v(x)= \frac{\pi}{2} – 2\theta = \frac{\pi}{2} – 2\cos^{-1}(x). \] Differentiating \(v(x)\) with respect to \(x\) gives: \[ v'(x)= -2\frac{d}{dx}\cos^{-1}(x)= -2\left(-\frac{1}{\sqrt{1-x^{2}}}\right)= \frac{2}{\sqrt{1-x^{2}}}. \] Therefore, the derivative of \(u\) with respect to \(v\) is: \[ \frac{du}{dv}= \frac{u'(x)}{v'(x)}= \frac{-\frac{1}{\sqrt{1-x^{2}}}}{\frac{2}{\sqrt{1-x^{2}}}}= -\frac{1}{2}. \]

Solution: We start by letting \[ u(x)= \tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right) \quad\text{and}\quad v(x)= \sin^{-1} x. \] Notice that by the half-angle formula, we have \[ \tan\frac{v}{2}= \frac{\sin v}{1+\cos v}. \] Since \( v=\sin^{-1} x \) implies \(\sin v=x\) and \(\cos v=\sqrt{1-x^2}\), it follows that \[ \tan\frac{v}{2}= \frac{x}{1+\sqrt{1-x^2}}. \] Hence, \[ u(x)= \tan^{-1}\left(\tan\frac{v}{2}\right)= \frac{v}{2}. \] Differentiating \( u \) with respect to \( v \) yields \[ \frac{du}{dv}= \frac{1}{2}. \] This result is independent of \( x \), which completes the proof.

Solution: Let \[ y=x^{x}. \] Express \( y \) in exponential form: \[ y=e^{x \log x}. \] Define \[ u=x \log x. \] Then, \[ y=e^{u}. \] Differentiating \( y \) with respect to \( u \) gives: \[ \frac{dy}{du}=e^{u}. \] Since \( e^{u}=e^{x \log x}=x^{x} \), the derivative of \( y \) with respect to \( x \log x \) is: \[ \frac{dy}{d(x \log x)}=x^{x}. \]