Question

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

Answer 1

Substitute g in for x to find f(g) from f(x), and recall `d/dx f(g(x))= (df)/(dg) (dg)/dx`. See explanation.

Explanation:

Recall that the chain rule states that if `y=f(g(x)), dy/dx =(df)/(dg)*(dg)/(dx)`

We have `f(x) =-e^x, g(x) = 1/sqrt(1-x)`

(This step is unnecessary, but in case the student is curious, to find f(g) you substitute g in for x in f(x), making `f(g) = -e^g`

We know that `d/(du) ke^u=ke^u`, and `d/dx(1/(1-x)^(1/2)) = 1/2 * 1/(1-x)^(3/2) = 1/(2(1-x)^(3/2))`.
Thus...

`dy/dx = (df)/(dg) (dg)/(dx)= 1/(2(1-x)^(3/2)) *- e^(1/sqrt(1-x)) = (-e^(1/sqrt(1-x)))/(2(1-x)^(3/2))`

Answer 2

`d/dx f(g(x)) = - (e^(sqrt(1-x)))/(2(1-x)^(3/2))`

Explanation:

`f(x) = -e^x and g(x) = 1/sqrt(1-x)`

Replace `x` in `f(x)` with g(x)#

`-> f(g(x)) = -e^(1/sqrt(i-x))`

`= -e^((1-x)^(-1/2))`

Apply the chain rule and standard differential

`d/dx f(g(x)) = -e^((1-x)^(-1/2)) * d/dx (1-x)^(-1/2)`

Apply the chain rule and power rule

`d/dx f(g(x)) = -e^((1-x)^(-1/2)) * (-1/2)(1-x)^(-3/2)* d/dx (1-x)`

`= -e^((1-x)^(-1/2)) * (-1/2)(1-x)^(-3/2)* (-1)`

`= -e^(1/sqrt(1-x)) * 1/(2(1-x)^(3/2))`

`= - (e^(sqrt(1-x)))/(2(1-x)^(3/2))`