Question

If `f(x)= - e^x` and `g(x) = 5 x`, how do you differentiate `f(g(x))` using the chain rule?

Answer 1

`-5e^(5x)`

Explanation:

`"to evaluate "f(g(x)" substitute "g(x)" into "f(x)`

`rArrf(g(x)=f(5x)=-e^(5x)`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`d/dx[f(g(x))]`

`=-e^(5x)xxd/dx(5x)=-5e^(5x)`

Answer 2

`d/dx(f(g(x)))=-5e^(5x)`

Explanation:

Given `f(x)=-e^x` and `g(x)=5x`

So,

`f(g(x))=-e^(g(x))`

As `g(x)=5x`

Therefore,

`f(g(x))=-e^(5x)`

Now differentiate both sides with respect to `x` using the chain rule.

The chain rule says that `->`

`(U(V(x)))'=U'(V(x)) xx V'(x)`

Now back to the question `->`

`f(g(x))=-e^(5x)`

`d/dx(f(g(x)))=d/dx-e^(5x)`

`d/dx(f(g(x)))=-e^(5x) xx d/dx 5x`

`d/dx(f(g(x)))=-e^(5x) xx 5`

Therefore,

`d/dx(f(g(x)))=-5e^(5x)`