# If f(x)= - x^2 + x  and g(x) = sqrtx , how do you differentiate f(g(x))  using the chain rule?

Sep 4, 2016

$\frac{1}{2 \sqrt{x}} - 1$

#### Explanation:

We have: $f \left(x\right) = - {x}^{2} + x$ and $g \left(x\right) = \sqrt{x}$

First, let's evaluate $f \left(g \left(x\right)\right)$:

$\implies f \left(g \left(x\right)\right) = f \left(\sqrt{x}\right)$

$\implies f \left(g \left(x\right)\right) = - {\left(\sqrt{x}\right)}^{2} + \left(\sqrt{x}\right)$

$\implies f \left(g \left(x\right)\right) = \sqrt{x} - x$

Now, let's differentiate this expression:

$\implies \frac{d}{\mathrm{dx}} \left(\sqrt{x} - x\right) = \frac{1}{2} {x}^{- \frac{1}{2}} - {x}^{0}$

$\implies \frac{d}{\mathrm{dx}} \left(\sqrt{x} - x\right) = \frac{1}{2 \sqrt{x}} - 1$