CUET Preparation Today
Differentiate the functions $\sin(\log x), x > 0$ with respect to $x$. |
$\frac{\sin(\log x)}{x}$ $\cos(\log x)$ $\frac{\cos(\log x)}{x}$ $-\frac{\cos(\log x)}{x}$ |
$\frac{\cos(\log x)}{x}$ |
The correct answer is Option (3) → $\frac{\cos(\log x)}{x}$ ## Let $y = \sin(\log x)$. Using chain rule, we have $\frac{dy}{dx} = \cos(\log x) \cdot \frac{d}{dx}(\log x) = \frac{\cos(\log x)}{x}$ |
