# If f(x)= 2 x^2 + x  and g(x) = sqrtx + 1 , how do you differentiate f(g(x))  using the chain rule?

Feb 23, 2016

$f ' \left(g \left(x\right)\right) = 2 + \frac{5}{2 \sqrt{x}}$

#### Explanation:

Find f(g(x)) $= f \left(\sqrt{x} + 1\right)$

substitute $x = \sqrt{x} + 1 \text{ in f(x) to obtain }$

f(g(x)) = $2 {\left(\sqrt{x} + 1\right)}^{2} + \left(\sqrt{x} + 1\right)$

distribute ${\left(\sqrt{x} + 1\right)}^{2} = x + 2 \sqrt{x} + 1$

$\Rightarrow f \left(g \left(x\right)\right) = 2 \left(x + 2 \sqrt{x} + 1\right) + \sqrt{x} + 1$

$= 2 x + 4 \sqrt{x} + 2 + \sqrt{x} + 1 = 2 x + 5 \sqrt{x} + 3$

and writing $2 x + 5 \sqrt{x} + 3 \text{ as } 2 x + 5 {x}^{\frac{1}{2}} + 3$

now differentiate using$\textcolor{b l u e}{\text{ power rule }}$

ie $\frac{d}{\mathrm{dx}} \left(a {x}^{n}\right) = n a {x}^{n - 1} \text{ term by term }$

$\Rightarrow f ' \left(g \left(x\right)\right) = 2 + \frac{5}{2} {x}^{- \frac{1}{2}} = 2 + \frac{5}{2 \sqrt{x}}$