1. ## Y=X^(2)e^(cos2x) Find Y'

Find Y' when Y= X^(2)e^(cos2x)

$Y= X^2e^{cos2x}$

These are the steps my teacher wrote down on the board. She used the Product rule.

a) Y'= [{X^(2)}' *{e^(cos2x)}] + [{X^(2)} *{e^(cos2x)}']

b) Y' = 2Xe^(cos2x) + X^(2)e^(cos2x) * [1-2sin(2x)]

c) Y' = 2Xe^(cos2x)[1-2sin(2x)]

So my questions are.
1) Where did that 1 come from?
2) How to go from step b to step c. Especifically why did X^(2)e^(cos2x) disapear in the final answer.

2. Originally Posted by Brazuca
Find Y' when Y= X^(2)e^(cos2x)

$Y= X^2e^{cos2x}$

These are the steps my teacher wrote down on the board. She used the Product rule.

a) Y'= [{X^(2)}' *{e^(cos2x)}] + [{X^(2)} *{e^(cos2x)}']

b) Y' = 2Xe^(cos2x) + X^(2)e^(cos2x) * [1-2sin(2x)]

c) Y' = 2Xe^(cos2x)[1-2sin(2x)]

So my questions are.
1) Where did that 1 come from?
2) How to go from step b to step c. Especifically why did X^(2)e^(cos2x) disapear in the final answer.
correction ...

$x^2e^{\cos(2x)}$

$2xe^{\cos(2x)} + x^2e^{\cos(2x)} [-2\sin(2x)]$

factor out $2xe^{\cos(2x)}$ from both terms ...

$2xe^{\cos(2x)}[1 - x\sin(2x)]$

3. Wait then did my teacher do this one wrong?

Also I don't understand how you factored out $
2xe^{\cos(2x)}
$
from both terms.

4. Originally Posted by Brazuca
Wait then did my teacher do this one wrong?

can't say ... I wasn't there. did you copy it correctly?

Also I don't understand how you factored out $
2xe^{\cos(2x)}
$
from both terms.
$2xe^{\cos(2x)} + x^2 e^{\cos(2x)}[-2\sin(2x)] =$

$\textcolor{red}{2xe^{\cos(2x)}} - \textcolor{red}{2x e^{\cos(2x)}}[x\sin(2x)]$

factor out $\textcolor{red}{2xe^{\cos(2x)}}$ ...

$\textcolor{red}{2xe^{\cos(2x)}}[1 - x\sin(2x)]$

5. Oh I understand it now. I just had to seperate the Xs from the Es in order for it all to click.