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

Answer 1

Answer: (A)

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

Since, $y=x^{\sin x}+\sqrt{x}$

Let $y_1=x^{\sin x}$ and $y_2=\sqrt{x}$

Taking log on both sides, we get

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

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

$\frac{1}{y_1}.\frac{{dy}_1}{dx}=\cos x\log x+\frac{1}{x}\sin x$

$\Rightarrow \frac{{dy}_1}{dx}=x^{\sin x}[\cos x\log x+\frac{1}{x}\sin x]$

$\therefore {\left(\frac{{dy}_1}{dx}\right)}_{x=\frac{\pi }{2}}={\left(\frac{\pi }{2}\right)}^{\sin \frac{\pi }{2}}[\cos \frac{\pi }{2}\log \frac{\pi }{2}+\frac{2}{\pi }\sin \frac{\pi }{2}]$

$=\frac{\pi }{2}\times \frac{2}{\pi }=1$

Now, $y_2=\sqrt{x}$

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

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

$\therefore {\left(\frac{{dy}_2}{dx}\right)}_{{}_{x=\frac{\pi }{2}}}=\frac{1}{2\sqrt{\frac{\pi }{2}}}=\sqrt{\frac{1}{2\pi }}$

Since, $y=y_1+y_2$

$\therefore \frac{dy}{dx}=\frac{{dy}_1}{dx}+\frac{{dy}_2}{dx}$

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