Question

Derivative of $f\left(x\right)=x\sin \left(\sqrt{x}\right)$ is

• A. $\frac{\cos \left(\sqrt{x}\right)}{2\sqrt{x}}+\sin \left(\sqrt{x}\right)$
• B. $\frac{\sqrt{x}.\cos \left(\sqrt{x}\right)}{2}-\sin \left(\sqrt{x}\right)$
• C. $\frac{\sqrt{x}.\cos \left(\sqrt{x}\right)}{2}+\sin \left(\sqrt{x}\right)$
• D. $\frac{\cos \left(\sqrt{x}\right)}{2\sqrt{x}}-\sin \left(\sqrt{x}\right)$

Answer 1

The correct option is C $\frac{\sqrt{x}.\cos \left(\sqrt{x}\right)}{2}+\sin \left(\sqrt{x}\right)$

Given, $f\left(x\right)=x\sin \left(\sqrt{x}\right)$
So, derivative of the given function is
$\frac{df\left(x\right)}{dx}=\frac{d\left(x\sin \right(\sqrt{x}\left)\right)}{dx}$
Using product and chain rule together we get
$x\frac{d\sin \left(\sqrt{x}\right)}{dx}+\sin \left(\sqrt{x}\right)\frac{dx}{dx}$
$\Rightarrow x[\frac{d\sin \left(\sqrt{x}\right)}{d\sqrt{x}}\times \frac{d\sqrt{x}}{dx}]+\sin \left(\sqrt{x}\right)$
$\Rightarrow x\left(\frac{\cos \left(\sqrt{x}\right)}{2\sqrt{x}}\right)+\sin \left(\sqrt{x}\right)$
$\Rightarrow \frac{\sqrt{x}.\cos \left(\sqrt{x}\right)}{2}+\sin \left(\sqrt{x}\right)$