The differential coefficient of a^x + log x c | vedclass

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Question

(A) Let $y = a^x + \log x \cdot \sin x$. Differentiating with respect to $x$,we apply the sum rule and the product rule: $\frac{dy}{dx} = \frac{d}{dx}

Vedclass · Accepted Answer

$a^x \log_e a + \frac{\sin x}{x} + \log x \cdot \cos x$

Vedclass · Answer

(A) Let $y = a^x + \log x \cdot \sin x$. Differentiating with respect to $x$,we apply the sum rule and the product rule: $\frac{dy}{dx} = \frac{d}{dx}(a^x) + \frac{d}{dx}(\log x \cdot \sin x)$. Using the standard derivative formula $\frac{d}{dx}(a^x) = a^x \log_e a$ and the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = a^x \log_e a + \left( \log x \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{d}{dx}(\log x) \right)$. $\frac{dy}{dx} = a^x \log_e a + \log x \cdot \cos x + \sin x \cdot \frac{1}{x}$. Thus,$\frac{dy}{dx} = a^x \log_e a + \frac{\sin x}{x} + \log x \cdot \cos x$. Therefore,the correct option is $A$.

The differential coefficient of $a^x + \log x \cdot \sin x$ is

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