Question

Find the derivative of:
y=x^2 e^-3x

Answer 1

You need to use the product rule here

Answer 2

\[(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\]

Answer 3

just treat f(x)=x^2 and g(x)=e^-3x

Answer 4

Calculate it and post your answer and I will let you know if you did it correctly

Answer 5

Okay so when we apply the power rule would it be like this:
(2x)(e^-3x)+(x^2)(e^-3x)

Answer 6

Or would the derivative of g(x) be: -3e^x?

Answer 7

You are almost correct.... except when you take the derivative:
\[\frac{d}{dx}e^u=e^u\frac{d}{dx}u=e^uu'\]

Answer 8

notice how the exponential remains unchanged

Answer 9

So the first term ((2x)(e^-3x) is correct but the second needs work

Answer 10

Oh okay. So for the second term would e^u be x^2 and u would be e^-3x?

Answer 11

No no e^u was meant to stand in for the exponential factor... the term where you take the derivative of x^2 was the first term (which you did correctly)... I was focusing on how to take the derivative of the exponential factor in the second term properly

Answer 12

Once you correctly differentiate the exponential factor you multiply it by x^2 (unchanged) and that is the second term in the sum

Answer 13

Oh okay. So e^-3x differentiate would be -3e^3x
So the second part overall would be x^2(-3e^-3x)

Answer 14

You missed the minus sign in the first part of that post but your final answer is correct :D now put it all together

Answer 15

And then the final answer would be e^-3x(2x-3x^2) since we can factor out e^-3x

Answer 16

Ok cool! Thanks for the help! :)

Answer 17

Correct :D :D you can also factor out an additional x if you want

Answer 18

Your welcome!