# Thread: Derivative of Exponential Function

1. ## Derivative of Exponential Function

I'm still having a bit of trouble with the concept of the derivative of the exponential function (e^x).

"If h(z) = (z^2)(1+e^-z), determine h'(-1)."

How do I solve this?

2. ## ok

Originally Posted by Jeavus
I'm still having a bit of trouble with the concept of the derivative of the exponential function (e^x).

"If h(z) = (z^2)(1+e^-z), determine h'(-1)."

How do I solve this?
ok you have $h(x)=x^2(1+e^{-x})$...then applying the product rule we get $h'(x)=x^2\cdot{-e^{-x}}+2x(1+e^{-x})$

3. ## Maybe

this will be helpful $\frac{D[e^{u(x)}]}{dx}=e^{u(x)}\cdot{u'(x)}$...so for $e^{-x}$ we have $u(x)=-x,u'(x)=-1$....so using the formula I have you we have $\frac{D[e^{-x}]}{dx}=e^{-x}\cdot{-1}=-e^{-x}$ as indicated above

4. Originally Posted by Jeavus
I'm still having a bit of trouble with the concept of the derivative of the exponential function (e^x).

"If h(z) = (z^2)(1+e^-z), determine h'(-1)."

How do I solve this?
$h(z)=z^2+z^2e^{-z}$ taking the derivative

$\frac{dh}{dz}=2z+\underbrace{2ze^{-z}-z^2e^{-z}}_{productRule}$

$\frac{dh}{dz}|_{z=-1}=2(-1)+2(-1)e^{1}-(-1)^2e^{1}=-2-3e$

5. Thanks you two!

6. ## Haha

Originally Posted by Jeavus
Thanks you two!
Its not as though I have an english term paper due for English 11 tomorrow...haha xD