# If f(x)= sqrt(x-2  and g(x) = 1/x , what is f'(g(x)) ?

##### 1 Answer
Jul 10, 2016

-1/(2x^2sqrt(1/x-2)

#### Explanation:

Before differentiating we require to establish f(g(x))

Substitute x $= \frac{1}{x} \text{into f(x)}$

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

now differentiate using the $\textcolor{b l u e}{\text{chain rule combined with power rule}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$

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

and $g \left(x\right) = {x}^{-} 1 - 2 \Rightarrow g ' \left(x\right) = - 1 {x}^{- 2} = - {x}^{- 2}$
$\text{--------------------------------------------------------}$
Substitute these values into (A)

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

Expressing the answer with positive indices

rArrf'(g(x))=-1/(2x^2sqrt(1/x-2)