Question

y=e^x√x
And dy/dx

IIT question.
Topic: Differentiation

Answer 1

Answer:

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Answer 2

Answer:

$\large\boxed{\sf{{e}^{x} ( \sqrt{x} + \dfrac{1}{2 \sqrt{x} } )}}$

Explanation:

Given a function such that,

$y = {e}^{x} \sqrt{x}$

Differentiatiating both the sides, wrt x, we get,

$= > \dfrac{dy}{dx} = \dfrac{d}{dx} {e}^{x} \sqrt{x}$

But, we know that,

• $\dfrac{d}{dx} uv = u \dfrac{dv}{dx} + v \dfrac{du}{dx}$

Therefore, we will get,

$= > \dfrac{dy}{dx} = {e}^{x} \dfrac{d}{dx} \sqrt{x} + \sqrt{x} \dfrac{d}{dx} {e}^{x}$

But, we know that,

• $\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} }$
• $\dfrac{d}{dx} {e}^{x} = {e}^{x}$

Therefore, we will get,

$= > \dfrac{dy}{dx} = {e}^{x} \times \dfrac{1}{2 \sqrt{x} } + {e}^{x} \sqrt{x} \\ \\ = > \bold{ \frac{dy}{dx} = {e}^{x} ( \sqrt{x} + \dfrac{1}{2 \sqrt{x} } )}$

Well IIT won't ask such an easy derivative.