Differentiate $\left(1+\sqrt{x}\right)^{8}$ by using the chain rule.

Solution
Chain rule: $\displaystyle\frac{du}{dx}=\displaystyle\frac{du}{dv}\displaystyle\frac{dv}{dx}$

We must determine the individual components of the chain rule to apply it. Set $u= \left(1+\sqrt{x}\right)^{8}$ and $v=1+\sqrt{x}$ .

Therefore,
$u=v^8$ ,
$\displaystyle\frac{du}{dv}=8v^7$ ,

$\displaystyle\frac{dv}{dx}=\displaystyle\frac{d}{dx}x^{\frac{1}{2}}=\displaystyle\frac{1}{2}x^{\frac{1}{2}-1}=\displaystyle\frac{1}{2}x^{-\frac{1}{2}}=\displaystyle\frac{1}{2\sqrt{x}}$ .

Hence,
$\displaystyle\frac{du}{dx}=\displaystyle\frac{du}{dv}\displaystyle\frac{dv}{dx}=\left(8v^7\right)\left(\displaystyle\frac{1}{2\sqrt{x}}\right)=\displaystyle\frac{4\left(1+\sqrt{x}\right)^7}{\sqrt{x}}$