derivative of e

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Sophiesmark

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Nov 4, 2005
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This one is even harder.
g(x)=e^-(x^2/4)
find, g' and g"

I think g'= -1/4e^-(x^2/4)

Am I way off here, and how do I find the second derivative of this.
Sophie
 
This one is even harder.
g(x)=e^-(x^2/4)
find, g' and g"

I think g'= -1/4e^-(x^2/4)

Am I way off here, and how do I find the second derivative of this.
Sophie
Click to expand...

Use the Chain Rule:
g(x) = f(h(x)), where f(x)=e^x and h(x)={-(x^2)/4}, so that:
g'(x) = f'(h(x))*h'(x) = {e^((-x^2)/4)}*{-2*x/4} = -(x/2)*e^((-x^2)/4)

Repeat process to obtain g''(x), except you'll first need to use the Product
Rule before using the Chain Rule.

.
 
I just made a mess out of this one, how do I do both the product and chain rule on the same one?
 
I just made a mess out of this one, how do I do both the product and chain rule on the same one?
Click to expand...

Begin with:
g'(x) = -(x/2)*e^((-x^2)/4)
Then apply Product Rule to obtain g''(x):
g''(x) = {(d/dx)(-x/2)}*{e^((-x^2)/4)} + {-(x/2)}*{(d/dx)e^((-x^2)/4)}
Then evaluate derivatives (d/dx), where you'll need the Chain Rule on the
last (d/dx) term on the right side (only that (d/dx) term needs the Chain Rule).

.
 
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