Question

Find the derivative :$\sqrt{x+1}$

Answer 1

$\sqrt{x+1}\therefore f\left(x\right)=\frac{d}{dx}\left[\sqrt{x+1}\right]=\frac{1}{2}{(x+1)}^{\frac{1}{2}-1}.\frac{d}{dx}[x+1]=\frac{\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[1\right]}{2\sqrt{x+1}}=\frac{1+0}{2\sqrt{x+1}}=\frac{1}{2\sqrt{x+1}}again,f^{\prime }\left(x\right)=\frac{d}{dx}\left[\frac{1}{2\sqrt{x+1}}\right]=\frac{1}{2}.\frac{d}{dx}\left[\frac{1}{\sqrt{x+1}}\right]=\frac{(-\frac{1}{2}){(x+1)}^{-\frac{1}{2}-1}.\frac{d}{dx}[x+1]}{2}=\frac{-\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[1\right]}{4{(x+1)}^{\frac{3}{2}}}=\frac{-1}{4{(x+1)}^{\frac{3}{2}}}$