Easy
Differentiate the following with respect to $x:$ $\sin (\log x), x > 0$
Let $y = \sin (\log x).$
Using the chain rule,we have:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{d}{dx}(\log x)$
Since $\frac{d}{dx}(\log x) = \frac{1}{x},$ we get:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{1}{x} = \frac{\cos (\log x)}{x}$
Using the chain rule,we have:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{d}{dx}(\log x)$
Since $\frac{d}{dx}(\log x) = \frac{1}{x},$ we get:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{1}{x} = \frac{\cos (\log x)}{x}$
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