Find the second order derivative of the funct | vedclass

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Question

(A) Let $y = \sin (\log x)$.First,differentiate with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}[\sin (\log x)] = \cos (\log x)

Vedclass · Accepted Answer

$\frac{-[\sin (\log x)+\cos (\log x)]}{x^{2}}$

Vedclass · Answer

(A) Let $y = \sin (\log x)$.First,differentiate with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}[\sin (\log x)] = \cos (\log x) \cdot \frac{d}{dx}(\log x) = \frac{\cos (\log x)}{x}$.Now,find the second order derivative $\frac{d^{2}y}{dx^{2}}$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^{2}}$:$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left[\frac{\cos (\log x)}{x}\right]$$= \frac{x \cdot \frac{d}{dx}[\cos (\log x)] - \cos (\log x) \cdot \frac{d}{dx}(x)}{x^{2}}$$= \frac{x \cdot [-\sin (\log x) \cdot \frac{1}{x}] - \cos (\log x) \cdot 1}{x^{2}}$$= \frac{-\sin (\log x) - \cos (\log x)}{x^{2}}$$= \frac{-[\sin (\log x) + \cos (\log x)]}{x^{2}}$.

Find the second order derivative of the function $\sin (\log x)$.

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