Derivative of Exponential Function

**Jeavus** · April 10th 2008, 05:10 PM

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?

**Mathstud28** · April 10th 2008, 05:15 PM

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 $h'(x)=x^2\cdot{-e^{-x}}+2x(1+e^{-x})$

**Mathstud28** · April 10th 2008, 05:16 PM

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

**TheEmptySet** · April 10th 2008, 05:19 PM

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

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

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

**Jeavus** · April 10th 2008, 05:22 PM

Thanks you two!

**Mathstud28** · April 10th 2008, 05:23 PM

Originally Posted by Jeavus

Thanks you two!

Its not as though I have an english term paper due for English 11 tomorrow...haha xD

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ok

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