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)

Since, $y=x^{\sin x}+\sqrt{x}$
Let $y_1=x^{\sin x}$ and $y_2=\sqrt{x}$
$\because y_1=x^{\sin x}$
Taking log on both sides, we get
$\log y_1=\sin x\log x$
On differentiating w.r.t. $x$, we get
$\frac{1}{y_1}\cdot \frac{d{y}_1}{dx}=\cos x\log x+\frac{1}{x}\sin x$
$\Rightarrow \frac{d{y}_1}{dx}=x^{\sin x}\left[\cos x\log x+\frac{1}{x}\sin x\right]$
$\therefore {\left(\frac{d{y}_1}{dx}\right)}_{x=\frac{\pi }{2}}={\left(\frac{\pi }{2}\right)}^{\sin \frac{\pi }{2}}\left[\cos \frac{\pi }{2}\log \frac{\pi }{2}+\frac{2}{\pi }\sin \frac{\pi }{2}\right]$
$=\frac{\pi }{2}\times \frac{2}{\pi }=1$
Now, $y_2=\sqrt{x}$
On differentiating w.r.t. $x$, we get
$\frac{d{y}_2}{dx}=\frac{1}{2\sqrt{x}}$
$\therefore {\left(\frac{d{y}_2}{dx}\right)}_{x=\frac{\pi }{2}}=\frac{1}{2\sqrt{\frac{\pi }{2}}}=\sqrt{\frac{1}{2\pi }}$
Since, $ry=y_1+y_2$
$\therefore \frac{dy}{dx}=\frac{d{y}_1}{dx}+\frac{d{y}_2}{dx}$
$=1+\frac{1}{\sqrt{2\pi }}$