Question

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

Answer 1

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

Explanation:

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

So,

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

As `g(x)=sqrtx`

Therefore,

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

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^(sqrtx)`

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

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

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

Therefore,

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