Question

How do you differentiate `f(x)=e^(-x^2-2x+1)` using the chain rule?

Answer 1

`-2(x+1)*e^(-x^2-2x+1)`

Explanation:

The chain rule basically says `dy/dx =(dy/dz)*(dz\/dx)` where `y=f(x)`

So in this case let `z=-x^2-2x+1`

Then `y = e^z`

`dz/dx = -2x-2`

`dy/dz = e^z`

`dy/dx = (-2x-2)*e^(-x^2-2x+1) = -2(x+1)*e^(-x^2-2x+1)`

Answer 2

You can do it another way using the chain rule more indirectly, based on the properties of `e`.

`f'(x)=-2xe^(-x^2-2x+1)-2e^(-x^2-2x+1)`

Explanation:

The derivative of `e^x` is always the derivative of `x` multiplied by `e^x`, or

`d/dx e^x = x' * e^x`.

This is why the derivative of `e^x` is just `e^x` when `x` is a variable on its own rather than an embedded function, because the derivative of a single variable like `x` is just `1`, so

`d/dx e^x=1*e^x=e^x`.

However, when you have embedded functions, you have to do more complex differentials with `x`.

In the case above, you have `-x^2-2x+1` instead of `x`. Work out the derivative of this normally,

`d/dx (-x^2-2x+1)=-2x-2`

and multiply by `e` to its original power, to get the final answer of

`(-2x-2)e^(-x^2-2x+1)=-2xe^(-x^2-2x+1)-2e^(-x^2-2x+1)`