Question

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

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

Answer 1

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

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