Question

The value of $\frac{dy}{dx}$ at $x=\frac{\pi }{2}$, where $y$ is given by $y=x^{\sin x}+\sqrt{x}$, is

• A. $1+\frac{1}{\sqrt{2\mathrm{\pi }}}$
• B. $1$
• C. $\frac{1}{\sqrt{2\mathrm{\pi }}}$
• D. $1-\frac{1}{\sqrt{2\mathrm{\pi }}}$

Answer 1

The correct option is A

$1+\frac{1}{\sqrt{2\mathrm{\pi }}}$

Explanation for the correct option:

Step 1. Find the derivative

Given function: $y=x^{\sin x}+\sqrt{x}$

Let $y_1=x^{\sin x}$

Taking $\log$ on both sides, we get

$\log y_1=\sin x\log x$

On differentiating w.r.t. $x$, we get

$$\begin{gathered} \left(\frac{1}{y_1}\right)\left(\frac{d{y}_1}{dx}\right)=\cos x\log x+\frac{1}{x}\sin x \\ \Rightarrow \frac{d{y}_1}{dx}=x^{\sin x}\left[\cos x\log x+\frac{1}{x}\sin x\right] \end{gathered}$$

$$\begin{gathered} \therefore {\left(\frac{d{y}_1}{dx}\right)}_{x=\frac{\pi }{2}}={\left(\frac{\pi }{2}\right)}^{{\sin }^{\frac{\pi }{2}}}\left[\cos \frac{\pi }{2}\log \frac{\pi }{2}+\frac{\pi }{2}\sin \frac{\pi }{2}\right] \\ =\frac{\pi }{2}\times \frac{2}{\pi } \\ =1 \end{gathered}$$

Step 2. Find the derivative

Now let $y_2=\sqrt{x}$

On differentiating w.r.t. $x$, we get

$\frac{d{y}_2}{dx}=\frac{1}{2\sqrt{x}}$

$$\begin{gathered} \therefore {\left(\frac{d{y}_2}{dx}\right)}_{x=\frac{\pi }{2}}=\frac{1}{2\sqrt{{\displaystyle \frac{\pi }{2}}}} \\ =\sqrt{\frac{1}{2\pi }} \end{gathered}$$

Step 3. Find the value of the derivative of the function

Since, $y=y_1+y_2$

$$\begin{gathered} \therefore \frac{dy}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx} \\ =1+\frac{1}{\sqrt{2\pi }} \end{gathered}$$

Hence, option (A) is the correct answer.