# If f(x)= 1/x  and g(x) = 1/x , how do you differentiate f'(g(x))  using the chain rule?

Mar 25, 2016

(df(g(x)))/(dx)=color(red)(-x^2)*color(blue)((-1/x^2)= 1

#### Explanation:

Given: $f \left(x\right) = \frac{1}{x}$ and $g \left(x\right) = \frac{1}{x}$
Required: $\frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = f ' \left(g \left(x\right)\right)$
Definition and Principles - Chain Rule:
$\frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = \textcolor{red}{\frac{\mathrm{df} \left(g\right)}{\mathrm{dg}}} \cdot \textcolor{b l u e}{\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}}$
$\textcolor{red}{\frac{\mathrm{df} \left(g\right)}{\mathrm{dg}}} = - \frac{1}{g} ^ 2$ but $g = \frac{1}{x}$, thus
$\textcolor{red}{\frac{\mathrm{df} \left(g\right)}{\mathrm{dg}}} = - \frac{1}{\frac{1}{{x}^{2}}} = - {x}^{2}$
$\textcolor{b l u e}{\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}} = - \frac{1}{x} ^ 2$ thus
(df(g(x)))/(dx)=color(red)(-x^2)*color(blue)((-1/x^2)= 1