Limits and Derivatives — MATHS STD 11 Science — Question

4. 0

Solution:

$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$

$=\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}$

Differentiate both the sides with respect to x, we get

$\frac{1}{2}\text{x}^{\frac{1}{2}-1}+\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}$

$\frac{1}{2}\text{x}^{-\frac{1}{2}}-\Big(\frac{1}{2}\Big)\text{x}^{-\frac{3}{2}}$

Putting x = 1, we get

$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1}=\frac{1}{2}\times1-\frac{1}{2}\times1=0$

Thus, $\frac{\text{dy}}{\text{dx}}$ at x = 1 is 0.