Question

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

Answer 1

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

Explanation:

Given: `f(x) = 1/x` and `g(x)=1/x`
Required: `(df(g(x)))/(dx)=f'(g(x))`
Definition and Principles - Chain Rule:
`(df(g(x)))/(dx)=color(red)((df(g))/(dg))*color(blue)((dg(x))/(dx))`
`color(red)((df(g))/(dg)) = -1/g^2` but `g=1/x`, thus
`color(red)((df(g))/(dg)) = -1/(1/(x^2))=-x^2`
`color(blue)((dg(x))/(dx))=-1/x^2` thus
`(df(g(x)))/(dx)=color(red)(-x^2)*color(blue)((-1/x^2)= 1`