Question

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

• A. $a^x\log a+\frac{\sin x}{x}+\log x\cos x$
• B. $a^x+\frac{\sin x}{x}+\log x\cos x$
• C. $a^x\log a+\frac{\cos x}{x}+\log x\sin x$
• D. None of these

Answer 1

The correct option is A

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

Explanation for the correct option:

Differentiating the given equation:

Let $y=a^x+\log x\sin x$

Differentiating w.r.t.$x$ by splitting into two terms :

Using Product Rule for $\log x\sin x$ and directly differentiating ${\mathrm{a}}^{\mathrm{x}}$ :

$\begin{array}{rcl}\frac{d\log x\sin \left(x\right)}{dx}& =& \left\{\frac{d\sin x}{dx}\right\}\log x+\sin x\frac{d\left(\log x\right)}{dx}\\ & =& \log x.\cos x+\frac{1}{x}\sin x\dots \dots \left(1\right)\\ \frac{{\mathrm{da}}^{\mathrm{x}}}{\mathrm{dx}}& =& {\mathrm{a}}^{\mathrm{x}}.\log \mathrm{a}\dots \dots \left(2\right)\end{array}$

The derivate of the sum is equal to the sum of their derivatives. So, adding equations $\left(1\right)$ and $\left(2\right)$ we get the required differential equation,

$\frac{dy}{dx}={\mathrm{a}}^{\mathrm{x}}.\log \mathrm{a}+\log x.\cos x+\frac{\sin x}{x}$

Hence, option (A) is the correct answer.