Ex 5.5 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: We differentiate each term individually: \[ \frac{d}{dx}(5 \tan x)=5\sec^2x, \] \[ \frac{d}{dx}(-3\sin x)=-3\cos x, \] \[ \frac{d}{dx}(4x^{\frac{3}{2}})=4\cdot\frac{3}{2}x^{\frac{1}{2}}=6\sqrt{x}. \] Combining these results, the derivative is: \[ f'(x)=5\sec^2x-3\cos x+6\sqrt{x}. \]

Solution: We differentiate term-by-term. For \( 3\cot x \): \[ \frac{d}{dx}(3\cot x) = 3 \left(-\csc^2 x\right) = -3\csc^2 x. \] For \( 6(1-2x)^{\frac{5}{3}} \): \[ \frac{d}{dx}\left[6(1-2x)^{\frac{5}{3}}\right] = 6 \cdot \frac{5}{3}(1-2x)^{\frac{2}{3}} \cdot (-2) = -20(1-2x)^{\frac{2}{3}}. \] Combining these results, the derivative is: \[ f'(x) = -3\csc^2 x – 20(1-2x)^{\frac{2}{3}}. \]

Solution: Using the product rule, where if \( f(x) = u(x)v(x) \) then \( f'(x)=u'(x)v(x)+u(x)v'(x) \), we set: \[ u(x)=x^3,\quad v(x)=\sin x. \] Then, \[ u'(x)=3x^2,\quad v'(x)=\cos x. \] Thus, \[ \frac{d}{dx}(x^3 \sin x)=3x^2\sin x + x^3\cos x. \]

Solution: We apply the product rule, where if \( f(x)=u(x)v(x) \), then \[ f'(x)=u'(x)v(x)+u(x)v'(x). \] Here, let: \[ u(x)=1+x^2,\quad v(x)=\cos x. \] We have: \[ u'(x)=2x,\quad v'(x)=-\sin x. \] Therefore, the derivative is: \[ \frac{d}{dx}\left[(1+x^2)\cos x\right]=2x\cos x – (1+x^2)\sin x. \]

Solution: We use the product rule where for \( f(x) = u(x)v(x) \), \[ f'(x)=u'(x)v(x)+u(x)v'(x). \] Let: \[ u(x)=x^3 \quad \text{and} \quad v(x)=\csc x. \] Then, \[ u'(x)=3x^2 \quad \text{and} \quad v'(x)=-\csc x\cot x. \] Thus, the derivative is: \[ \frac{d}{dx}(x^3 \csc x)=3x^2\csc x – x^3\csc x\cot x. \]

Solution: Let \[ u(x)=1-2\tan x \quad \text{and} \quad v(x)=5+4\sin x. \] First, we differentiate each function: \[ u'(x)=-2\sec^2 x, \quad \text{and} \quad v'(x)=4\cos x. \] Applying the product rule: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x), \] we have: \[ \frac{d}{dx}\left[(1-2\tan x)(5+4\sin x)\right] = -2\sec^2 x\,(5+4\sin x) + (1-2\tan x)(4\cos x). \]

Solution: Let \[ u(x)=x+\sin x \quad \text{and} \quad v(x)=x+\cos x. \] Then, \[ u'(x)=1+\cos x \quad \text{and} \quad v'(x)=1-\sin x. \] Using the quotient rule, \[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}, \] we have: \[ \frac{d}{dx}\left(\frac{x+\sin x}{x+\cos x}\right)=\frac{(1+\cos x)(x+\cos x)-(x+\sin x)(1-\sin x)}{(x+\cos x)^2}. \] Expanding the numerator: \[ (1+\cos x)(x+\cos x)= x+ x\cos x + \cos x+\cos^2 x, \] \[ (x+\sin x)(1-\sin x)= x – x\sin x + \sin x-\sin^2 x. \] Thus, the numerator becomes: \[ \Bigl[x+ x\cos x + \cos x+\cos^2 x\Bigr]-\Bigl[x – x\sin x + \sin x-\sin^2 x\Bigr], \] which simplifies to: \[ x\cos x + x\sin x + (\cos x-\sin x) + (\cos^2 x+\sin^2 x). \] Since \(\cos^2 x+\sin^2 x=1\), the numerator is: \[ x(\cos x+\sin x)+(\cos x-\sin x)+1. \] Therefore, the derivative is: \[ \frac{d}{dx}\left(\frac{x+\sin x}{x+\cos x}\right)=\frac{x(\cos x+\sin x)+(\cos x-\sin x)+1}{(x+\cos x)^2}. \]

Solution: Let \[ u(x)=\sin x+\cos x \quad \text{and} \quad v(x)=\sin x-\cos x. \] Then, we have: \[ u'(x)=\cos x-\sin x \quad \text{and} \quad v'(x)=\cos x+\sin x. \] Using the quotient rule, \[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}, \] we compute the numerator: \[ u'(x)v(x)=(\cos x-\sin x)(\sin x-\cos x)=-\left(\cos x-\sin x\right)^2, \] \[ u(x)v'(x)=(\sin x+\cos x)(\cos x+\sin x)=(\sin x+\cos x)^2. \] Thus, the numerator becomes: \[ -\left(\cos x-\sin x\right)^2-(\sin x+\cos x)^2. \] Note that: \[ (\cos x-\sin x)^2=\cos^2 x-2\sin x\cos x+\sin^2 x=1-2\sin x\cos x, \] \[ (\sin x+\cos x)^2=\sin^2 x+2\sin x\cos x+\cos^2 x=1+2\sin x\cos x. \] Therefore, the numerator simplifies to: \[ -\left[1-2\sin x\cos x\right]-\left[1+2\sin x\cos x\right]=-1+2\sin x\cos x-1-2\sin x\cos x=-2. \] Hence, the derivative is: \[ \frac{d}{dx}\left(\frac{\sin x+\cos x}{\sin x-\cos x}\right)=\frac{-2}{(\sin x-\cos x)^2}. \]

Solution: Let \[ u(x)=\sec x-1 \quad \text{and} \quad v(x)=\sec x+1. \] We compute the derivatives: \[ u'(x)=\sec x\tan x, \quad v'(x)=\sec x\tan x. \] Using the quotient rule, \[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}, \] we substitute: \[ \frac{d}{dx}\left(\frac{\sec x-1}{\sec x+1}\right)=\frac{\sec x\tan x (\sec x+1)-(\sec x-1)\sec x\tan x}{(\sec x+1)^2}. \] Factor out \(\sec x\tan x\): \[ =\frac{\sec x\tan x\left[(\sec x+1)-(\sec x-1)\right]}{(\sec x+1)^2}. \] Simplifying the bracket: \[ (\sec x+1)-(\sec x-1)=2, \] so the derivative becomes: \[ \frac{d}{dx}\left(\frac{\sec x-1}{\sec x+1}\right)=\frac{2\sec x\tan x}{(\sec x+1)^2}. \]

Solution: Let \[ u(x)=a+\sin x \quad \text{and} \quad v(x)=1+a\sin x. \] Then, the derivatives are: \[ u'(x)=\cos x,\quad v'(x)=a\cos x. \] Using the quotient rule, \[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}, \] we have: \[ \frac{d}{dx}\left(\frac{a+\sin x}{1+a\sin x}\right)=\frac{\cos x(1+a\sin x)- (a+\sin x)(a\cos x)}{(1+a\sin x)^2}. \] Factor \(\cos x\) in the numerator: \[ =\frac{\cos x\left[(1+a\sin x)-a(a+\sin x)\right]}{(1+a\sin x)^2}. \] Simplify inside the bracket: \[ (1+a\sin x)-a(a+\sin x)=1+a\sin x-a^2-a\sin x=1-a^2. \] Thus, the derivative becomes: \[ \frac{d}{dx}\left(\frac{a+\sin x}{1+a\sin x}\right)=\frac{(1-a^2)\cos x}{(1+a\sin x)^2}. \]

Solution: Let \[ u(x) = x \sin x \quad \text{and} \quad v(x) = 1 + \cos x. \] Now, differentiate both \( u(x) \) and \( v(x) \): \[ u'(x) = \sin x + x \cos x, \quad v'(x) = -\sin x. \] Using the quotient rule: \[ \frac{d}{dx}\left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) – u(x)v'(x)}{[v(x)]^2}, \] Substituting the values: \[ = \frac{(\sin x + x \cos x)(1 + \cos x) – (x \sin x)(-\sin x)}{(1 + \cos x)^2}. \] Simplify the expression: \[ = \frac{(\sin x + x \cos x)(1 + \cos x) + x \sin^2 x}{(1 + \cos x)^2}. \]

Solution: Let \[ u(x) = \sin x – x \cos x, \quad v(x) = x \sin x + \cos x. \] First, compute the derivatives: \[ u'(x) = \cos x – (\cos x – x \sin x) = x \sin x, \] (Using product rule for \( x \cos x \): \( \frac{d}{dx}(x \cos x) = \cos x – x \sin x \)) \[ v'(x) = \sin x + x \cos x – \sin x = x \cos x. \] (Using product rule for \( x \sin x \): \( \frac{d}{dx}(x \sin x) = \sin x + x \cos x \)) Now apply the quotient rule: \[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) – u(x)v'(x)}{[v(x)]^2}. \] Substituting the values: \[ = \frac{(x \sin x)(x \sin x + \cos x) – (\sin x – x \cos x)(x \cos x)}{(x \sin x + \cos x)^2}. \]

Solution: Let \[ y = \sqrt{ \frac{1 + \cos x}{1 – \cos x} } \] Rewrite the square root in exponent form: \[ y = \left( \frac{1 + \cos x}{1 – \cos x} \right)^{1/2} \] Let \[ u = \frac{1 + \cos x}{1 – \cos x}, \quad \text{then} \quad y = u^{1/2} \] Using chain rule: \[ \frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx} \] Now, differentiate \( u \) using the quotient rule. Let \[ f(x) = 1 + \cos x, \quad f'(x) = -\sin x \] \[ g(x) = 1 – \cos x, \quad g'(x) = \sin x \] Now apply the quotient rule: \[ \frac{du}{dx} = \frac{f'(x)g(x) – f(x)g'(x)}{[g(x)]^2} \] Substituting: \[ \frac{du}{dx} = \frac{(-\sin x)(1 – \cos x) – (1 + \cos x)(\sin x)}{(1 – \cos x)^2} \] Simplify the numerator: \[ = -\sin x (1 – \cos x) – \sin x (1 + \cos x) \] \[ = -\sin x + \sin x \cos x – \sin x – \sin x \cos x \] \[ = -2 \sin x \] So, \[ \frac{du}{dx} = \frac{-2 \sin x}{(1 – \cos x)^2} \] Now plug back into the chain rule: \[ \frac{dy}{dx} = \frac{1}{2} \cdot \left( \frac{1 + \cos x}{1 – \cos x} \right)^{-1/2} \cdot \frac{-2 \sin x}{(1 – \cos x)^2} \] Cancel out 2: \[ \frac{dy}{dx} = \frac{ – \sin x }{ (1 – \cos x)^2 \cdot \sqrt{ \frac{1 + \cos x}{1 – \cos x} } } \] Convert the square root: \[ \sqrt{ \frac{1 + \cos x}{1 – \cos x} } = \frac{ \sqrt{1 + \cos x} }{ \sqrt{1 – \cos x} } \] So: \[ \frac{dy}{dx} = \frac{ -\sin x \cdot \sqrt{1 – \cos x} }{ (1 – \cos x)^2 \cdot \sqrt{1 + \cos x} } \] Combine powers: \[ \frac{dy}{dx} = \frac{ -\sin x }{ (1 – \cos x)^{3/2} (1 + \cos x)^{1/2} } \]

Solution: Let \[ y = \sqrt{ \frac{ \sec x + \tan x }{ \sec x – \tan x } } \] Rewrite the square root in exponent form: \[ y = \left( \frac{ \sec x + \tan x }{ \sec x – \tan x } \right)^{1/2} \] Let \[ u = \frac{ \sec x + \tan x }{ \sec x – \tan x }, \quad \text{then} \quad y = u^{1/2} \] Using chain rule: \[ \frac{dy}{dx} = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx} \] Now, differentiate \( u \) using the quotient rule. Let \[ f(x) = \sec x + \tan x, \quad f'(x) = \sec x \tan x + \sec^2 x \] \[ g(x) = \sec x – \tan x, \quad g'(x) = \sec x \tan x – \sec^2 x \] Now apply the quotient rule: \[ \frac{du}{dx} = \frac{f'(x)g(x) – f(x)g'(x)}{[g(x)]^2} \] Substituting everything: \[ \frac{du}{dx} = \frac{(\sec x \tan x + \sec^2 x)(\sec x – \tan x) – (\sec x + \tan x)(\sec x \tan x – \sec^2 x)}{(\sec x – \tan x)^2} \] Let’s simplify both terms in the numerator: **First term:** \[ (\sec x \tan x + \sec^2 x)(\sec x – \tan x) \] Distribute: \[ = \sec^2 x \cdot \sec x – \sec^2 x \cdot \tan x + \sec x \tan x \cdot \sec x – \sec x \tan x \cdot \tan x \] \[ = \sec^3 x – \sec^2 x \tan x + \sec^2 x \tan x – \sec x \tan^2 x \] \[ = \sec^3 x – \sec x \tan^2 x \] **Second term:** \[ (\sec x + \tan x)(\sec x \tan x – \sec^2 x) \] Distribute: \[ = \sec x \cdot \sec x \tan x – \sec x \cdot \sec^2 x + \tan x \cdot \sec x \tan x – \tan x \cdot \sec^2 x \] \[ = \sec^2 x \tan x – \sec^3 x + \sec x \tan^2 x – \sec^2 x \tan x \] \[ = -\sec^3 x + \sec x \tan^2 x \] Now subtract second term from first term: \[ \left( \sec^3 x – \sec x \tan^2 x \right) – \left( -\sec^3 x + \sec x \tan^2 x \right) \] \[ = \sec^3 x – \sec x \tan^2 x + \sec^3 x – \sec x \tan^2 x \] \[ = 2\sec^3 x – 2\sec x \tan^2 x \] So, \[ \frac{du}{dx} = \frac{2\sec^3 x – 2\sec x \tan^2 x}{(\sec x – \tan x)^2} \] Factor out \( 2\sec x \): \[ \frac{du}{dx} = \frac{2\sec x (\sec^2 x – \tan^2 x)}{(\sec x – \tan x)^2} \] Using identity: \[ \sec^2 x – \tan^2 x = 1 \] So, \[ \frac{du}{dx} = \frac{2\sec x}{(\sec x – \tan x)^2} \] Now plug back into the derivative of \( y \): \[ \frac{dy}{dx} = \frac{1}{2} \cdot \left( \frac{ \sec x + \tan x }{ \sec x – \tan x } \right)^{-1/2} \cdot \frac{2\sec x}{(\sec x – \tan x)^2} \] Cancel out 2: \[ \frac{dy}{dx} = \frac{ \sec x }{ (\sec x – \tan x)^2 \cdot \sqrt{ \frac{ \sec x + \tan x }{ \sec x – \tan x } } } \] Convert the square root: \[ \sqrt{ \frac{ \sec x + \tan x }{ \sec x – \tan x } } = \frac{ \sqrt{ \sec x + \tan x } }{ \sqrt{ \sec x – \tan x } } \] Final expression: \[ \frac{dy}{dx} = \frac{ \sec x \cdot \sqrt{ \sec x – \tan x } }{ (\sec x – \tan x)^2 \cdot \sqrt{ \sec x + \tan x } } \] Simplify: \[ \frac{dy}{dx} = \frac{ \sec x }{ (\sec x – \tan x)^{3/2} (\sec x + \tan x)^{1/2} } \] Multiply numerator and denominator by \( \sqrt{\sec x + \tan x} \) to simplify: \[ \frac{dy}{dx} = (\sec x + \tan x) \sec x \]

Solution: Given: \[ y = \frac{2 – 3 \cos x}{\sin x} \] To find \( \frac{dy}{dx} \), we will use the **quotient rule**: Let \[ u = 2 – 3 \cos x \quad \text{and} \quad v = \sin x \] Then \[ u’ = 3 \sin x, \quad v’ = \cos x \] Using the quotient rule: \[ \frac{dy}{dx} = \frac{u’v – uv’}{v^2} = \frac{(3 \sin x)(\sin x) – (2 – 3 \cos x)(\cos x)}{\sin^2 x} \] Simplify the numerator: \[ 3 \sin^2 x – (2 \cos x – 3 \cos^2 x) = 3 \sin^2 x – 2 \cos x + 3 \cos^2 x \] So, \[ \frac{dy}{dx} = \frac{3 \sin^2 x + 3 \cos^2 x – 2 \cos x}{\sin^2 x} \] Since \( \sin^2 x + \cos^2 x = 1 \), we can write: \[ 3 \sin^2 x + 3 \cos^2 x = 3 \] Therefore, \[ \frac{dy}{dx} = \frac{3 – 2 \cos x}{\sin^2 x} \] Now, evaluate at \( x = \frac{\pi}{4} \): \[ \cos\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}, \quad \sin\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \Rightarrow \sin^2\left( \frac{\pi}{4} \right) = \frac{1}{2} \] Plug in the values: \[ \frac{dy}{dx}\bigg|_{x = \frac{\pi}{4}} = \frac{3 – 2 \cdot \frac{1}{\sqrt{2}}}{\frac{1}{2}} = 2 \cdot \left( 3 – \frac{2}{\sqrt{2}} \right) = 2 \cdot \left( 3 – \sqrt{2} \right) = 6 – 2\sqrt{2} \]

Solution: Given: \[ y = \left( \sin \frac{x}{2} + \cos \frac{x}{2} \right)^2 \] Using the identity: \[ \left( \sin \frac{x}{2} + \cos \frac{x}{2} \right)^2 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 + \sin x \] (Since \( 2 \sin \frac{x}{2} \cos \frac{x}{2} = \sin x \)) So, \[ y = 1 + \sin x \Rightarrow \frac{dy}{dx} = \cos x \] Now, evaluate at \( x = \frac{\pi}{6} \): \[ \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \]