The derivative of sin left( log left( frac{x | vedclass

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(C) Let $y = \sin \left( \log \left( \frac{x + 3}{x} \right) \right) = \sin (\log(x+3) - \log x)$.Applying the chain rule, $\frac{dy}{dx} = \cos \left

Vedclass · Accepted Answer

$\frac{-3}{x(x + 3)} \cos \left( \log \left( \frac{x + 3}{x} \right) \right)$

Vedclass · Answer

(C) Let $y = \sin \left( \log \left( \frac{x + 3}{x} \right) \right) = \sin (\log(x+3) - \log x)$.Applying the chain rule, $\frac{dy}{dx} = \cos \left( \log \left( \frac{x + 3}{x} \right) \right) \cdot \frac{d}{dx} \left( \log \left( \frac{x + 3}{x} \right) \right)$.Now, $\frac{d}{dx} \left( \log \left( \frac{x + 3}{x} \right) \right) = \frac{x}{x+3} \cdot \frac{d}{dx} \left( 1 + \frac{3}{x} \right)$.$= \frac{x}{x+3} \cdot \left( -\frac{3}{x^2} \right) = -\frac{3}{x(x+3)}$.Thus, $\frac{dy}{dx} = \cos \left( \log \left( \frac{x + 3}{x} \right) \right) \cdot \left( -\frac{3}{x(x+3)} \right) = \frac{-3}{x(x+3)} \cos \left( \log \left( \frac{x + 3}{x} \right) \right)$.

The derivative of $\sin \left( \log \left( \frac{x + 3}{x} \right) \right)$ with respect to $x$ is:

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