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
$\frac{1}{y_{1}} \cdot \frac{d y_{1}}{d x}=\cos x \log x+\frac{1}{x} \sin x$
$\Rightarrow \frac{d y_{1}}{d x}=x^{\sin x}\left[\cos x \log x+\frac{1}{x} \sin x\right]$
$\therefore \left(\frac{d y_{1}}{d x}\right)_{x=\frac{\pi}{2}} =\left(\frac{\pi}{2}\right)^{\sin \frac{\pi}{2}}\left[\cos \frac{\pi}{2} \log \frac{\pi}{2}+\frac{2}{\pi} \sin \frac{\pi}{2}\right]$
$=\frac{\pi}{2} \times \frac{2}{\pi}=1$
Now, $y_{2} =\sqrt{x}$
On differentiating w.r.t. $x$, we get
$\frac{d y_{2}}{d x} =\frac{1}{2 \sqrt{x}}$
$\therefore \left(\frac{d y_{2}}{d x}\right)_{x=\frac{\pi}{2}} =\frac{1}{2 \sqrt{\frac{\pi}{2}}}=\sqrt{\frac{1}{2 \pi}}$
Since, ${r}y =y_{1}+y_{2}$
$\therefore \frac{d y}{d x} =\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x}$
$=1+\frac{1}{\sqrt{2 \pi}}$