Class 12 Maths NCERT Solutions Chapter 5 Ex 5.4 – Exponential and Logarithmic Differentiation

Apply the quotient rule with u=e^x, v=\sin x: \dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^2}=\dfrac{e^x\sin x-e^x\cos x}{\sin^2x}.

Let y=e^{\sin^{-1}x}. By the chain rule: \dfrac{dy}{dx}=e^{\sin^{-1}x}\cdot\dfrac{d}{dx}\left(\sin^{-1}x\right).

Since \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}, substitute in.

Let y=e^{x^3}. By the chain rule: \dfrac{dy}{dx}=e^{x^3}\cdot\dfrac{d}{dx}(x^3)=e^{x^3}\cdot 3x^2.

This is a chain of three functions: u=e^{-x}, v=\tan^{-1}u, y=\sin v.

Differentiate outward: \dfrac{dy}{dv}=\cos v, \dfrac{dv}{du}=\dfrac{1}{1+u^2}, \dfrac{du}{dx}=-e^{-x}.

Multiply by the extended chain rule: \dfrac{dy}{dx}=\cos\left(\tan^{-1}e^{-x}\right)\cdot\dfrac{1}{1+e^{-2x}}\cdot(-e^{-x}).

Let y=\log(\cos e^x). By the chain rule: \dfrac{dy}{dx}=\dfrac{1}{\cos e^x}\cdot\dfrac{d}{dx}\left(\cos e^x\right).

\dfrac{d}{dx}\cos e^x=-\sin(e^x)\cdot e^x.

Substitute: \dfrac{dy}{dx}=\dfrac{-e^x\sin(e^x)}{\cos(e^x)}.

This is a sum of five terms e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}; differentiate each term separately using the chain rule.

\dfrac{d}{dx}e^x=e^x, \dfrac{d}{dx}e^{x^2}=2xe^{x^2}, \dfrac{d}{dx}e^{x^3}=3x^2e^{x^3}, \dfrac{d}{dx}e^{x^4}=4x^3e^{x^4}, \dfrac{d}{dx}e^{x^5}=5x^4e^{x^5}.

Let y=\left(e^{\sqrt{x}}\right)^{1/2}. By the chain rule: \dfrac{dy}{dx}=\dfrac{1}{2}\left(e^{\sqrt{x}}\right)^{-1/2}\cdot e^{\sqrt{x}}\cdot\dfrac{d}{dx}\left(\sqrt{x}\right).

Simplify \left(e^{\sqrt{x}}\right)^{-1/2}\cdot e^{\sqrt{x}}=e^{\sqrt{x}/2}, and \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}.

Combine: \dfrac{dy}{dx}=\dfrac{1}{2}\cdot e^{\sqrt{x}/2}\cdot\dfrac{1}{2\sqrt{x}}.

Let y=\log(\log x). By the chain rule: \dfrac{dy}{dx}=\dfrac{1}{\log x}\cdot\dfrac{d}{dx}(\log x)=\dfrac{1}{\log x}\cdot\dfrac{1}{x}.

Apply the quotient rule with u=\cos x, v=\log x: \dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^2}=\dfrac{-\sin x\log x-\cos x\cdot\frac{1}{x}}{(\log x)^2}.

Multiply numerator and denominator by x to clear the fraction inside: \dfrac{dy}{dx}=\dfrac{-x\sin x\log x-\cos x}{x(\log x)^2}.

Let y=\cos(\log x+e^x). By the chain rule: \dfrac{dy}{dx}=-\sin(\log x+e^x)\cdot\dfrac{d}{dx}(\log x+e^x).

\dfrac{d}{dx}(\log x+e^x)=\dfrac{1}{x}+e^x.