# Thread: Find the derivative: y=x^2e^-1/x?

1. ## Find the derivative: y=x^2e^-1/x?

can someone help me out with this?! step by step preferably, not just the answer, i want to be able to understand it!

2. ## Re: Find the derivative: y=x^2e^-1/x?

hi stargazer00

where are you stuck?

you should use product rule.

3. ## Re: Find the derivative: y=x^2e^-1/x?

i don't even know where to begin, if i see how to solve it, i feel like i can follow that model to solve problems similar to it.

4. ## Re: Find the derivative: y=x^2e^-1/x?

Originally Posted by stargazer00
can someone help me out with this?! step by step preferably, not just the answer, i want to be able to understand it!
Use the product rule:
$\frac{d}{{dx}}\left( {x^2 e^{ - 1/x} } \right) = \frac{d}{{dx}}\left( {x^2 } \right)\left[ {e^{ - 1/x} } \right] + \left[ {x^2 }\right]\frac{d}{{dx}}\left( {e^{ - 1/x} } \right)$

5. ## Re: Find the derivative: y=x^2e^-1/x?

Originally Posted by Plato
Use the product rule:
$\frac{d}{{dx}}\left( {x^2 e^{ - 1/x} } \right) = \frac{d}{{dx}}\left( {x^2 } \right)\left[ {e^{ - 1/x} } \right] + \left[ {x^2 }\right]\frac{d}{{dx}}\left( {e^{ - 1/x} } \right)$
And the chain rule will be needed to get the derivative of e^(-1/x) ....

6. ## Re: Find the derivative: y=x^2e^-1/x?

using f'(g(x)) * g'(x)

y= (x^2)e^(-1/x)
f(x) = x^2
f'(x) = 2x

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

2(e^(-1/x)) *1/2e^((-1/x)-1)

now what?

7. ## Re: Find the derivative: y=x^2e^-1/x?

Originally Posted by stargazer00

using f'(g(x)) * g'(x)

y= (x^2)e^(-1/x)
f(x) = x^2
f'(x) = 2x

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

2(e^(-1/x)) *1/2e^((-1/x)-1) what is this? do you know how to take the derivative of an exponential function?

now what?
incorrect ... note that the product rule is the primary rule to be used here as shown in Plato's post.

8. ## Re: Find the derivative: y=x^2e^-1/x?

Stargazer... $\frac{d}{dx}(e^u) = [e^u][u^\prime]$. Does that help?