The derivative of sin x with respect to log x | vedclass

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(B) Let $u = \sin x$ and $v = \log x$. We need to find the derivative of $u$ with respect to $v$,which is $\frac{du}{dv}$. Using the chain rule,we hav

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$x \cos x$

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(B) Let $u = \sin x$ and $v = \log x$. We need to find the derivative of $u$ with respect to $v$,which is $\frac{du}{dv}$. Using the chain rule,we have $\frac{du}{dv} = \frac{du/dx}{dv/dx}$. First,find the derivative of $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$. Next,find the derivative of $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$. Now,substitute these into the formula: $\frac{du}{dv} = \frac{\cos x}{1/x} = x \cos x$.

The derivative of $\sin x$ with respect to $\log x$ is

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