Question

If y=2^root x then dy/dx is

Answer 1

Answer:

The answer is $\dfrac{\log 2}{2}(\dfrac{2^{\sqrt x}}{\sqrt x})$

Step-by-step explanation:

Given $y=2^{\sqrt x}$

Thus using chain rule of differentiation

$\dfrac{dy}{dx}=\dfrac{d(2^{\sqrt x})}{dx}=\dfrac{d(2^{\sqrt x})}{d(\sqrt x})}\times \dfrac{d(\sqrt x)}{dx}$

                      $=2^{\sqrt x}\log 2\times \dfrac{1}{2\sqrt x}=\dfrac{\log 2}{2}(\dfrac{2^{\sqrt x}}{\sqrt x})$

• Formula : $\dfrac{d}{dx}a^x=a^x\log a$ and $\dfrac{d}{dx}(x^n)=nx^{n-1}$

Answer 2

Answer:

this is answer

Step-by-step explanation:

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