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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $y=\frac{\cos x}{1+\sin x}$,then
$(a)$ $\frac{dy}{dx}=\frac{-1}{1+\sin x}$
$(b)$ $\frac{dy}{dx}=\frac{1}{1+\sin x}$
$(c)$ $\frac{dy}{dx}=-\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$
$(d)$ $\frac{dy}{dx}=\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$

Derivative of $e^x$ with respect to $\sqrt{x}$ is

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to

If $f(x) = x \tan^{-1} x$,then $\lim_{x \rightarrow 1} \frac{f(x) - f(1)}{x - 1}$ is equal to

Differentiate the function with respect to $x$: $\sin^{3} x + \cos^{6} x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module