# How do you differentiate y=1/(x+sqrtx) using the chain or quotient rule?

Nov 28, 2016

I will show you how to solve this problem using the chain rule.

We can rewrite $y$ as $y = {\left(x + \sqrt{x}\right)}^{-} 1$.

We let $y = {u}^{-} 1$ and $u = x + \sqrt{x}$.

$\frac{\mathrm{dy}}{\mathrm{du}} = - 1 {u}^{-} 2 = - \frac{1}{u} ^ 2$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 1 + \frac{1}{2} {x}^{- \frac{1}{2}} = 1 + \frac{1}{2 {x}^{\frac{1}{2}}}$

By the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{u} ^ 2 \times \left(1 + \frac{1}{2 {x}^{\frac{1}{2}}}\right)$

dy/dx = (-2x^(1/2) + 1)/((x + sqrt(x))^2(2x^(1/2))

Hopefully this helps!