Question

Can someone teach me how to do this?



differentiate 2x^2e^(cos2x)

Answer 1

Use the product rule

Answer 2

I suggest factoring out a 2 first

Answer 3

Im just so lost, can you do it then teach me how you did it?

Answer 4

Okay man, her we go:
2[x^2e^(cos2x)]

Answer 5

ok

Answer 6

All I did here was factor out a 2

Answer 7

yes

Answer 8

so firstly, factor out all constants

Answer 9

Now we apply the product rule:
2[2xe^(cos2x)+x^2e^(cos2x)*(-sin2x)]

Answer 10

Where does the -sin come from?

Answer 11

It's from using the chian rule on e^(cos2x). The derivative of cos2x is -2sin2x

Answer 12

I forgot the 2 in the answer I gave you above it should be :

Now we apply the product rule:
2[2xe^(cos2x)+x^2e^(cos2x)*(-2sin2x)]

Answer 13

Remeber that if you have e^(u), then the derivative is u'e^(u)

Answer 14

ok I understand what you did now

Answer 15

so what next?

Answer 16

@LagrangeSon678 ?

Answer 17

That's it we're done, the only we can do is clean it up a bit if you want

Answer 18

Yes because in my textbook, the answer is :-4xe^(cos2x)*(xsin2x-1)

Answer 19

So how would we write it like how it is in the book?

Answer 20

Okay cool, so we have , 2[2xe^(cos2x)+x^2e^(cos2x)*(-2sin2x)]

So let's clean it up a little: 2[2xe^(cos2x)-2(sin2x)x^2e^(cos2x)]

Answer 21

ok, so you rearranged

Answer 22

Now if you notice there are a couple of terms that we can factor out,mainly inside the brackets there is a 2 and a e^(cos2x) that is common to both terms inside the brackets

Answer 23

yes

Answer 24

@LagrangeSon678 It still doesnt seem to be the same as the answer in the book

Answer 25

If I remove what you said we get 2[x-1(sin2x)x^2] which equals
2x-2(sin2x)2x^2

Answer 26

@LagrangeSon678 ?

Answer 27

@LagrangeSon678 Are you there?