Question

Differentiate
e^xsin(2x)

Answer 1

You need to use the chain rule and the product rule for this. This function is the product of e^x and sin(2x). Use the product rule to solve this and use chain rule to differentiate sin(2x)

Answer 2

could you please show me.

Answer 3

Do you know the product rule?

Answer 4

yes, I know both product and quotient rule but when I worked it out I did not get the right answer so I am asking for you to show me the details so i can see where I went wrong

Answer 5

Do you mind showing me your working?

Answer 6

\[e^xsin(2x) * (\sin(2x) + 2xcos(2x))\]

Answer 7

Ok. Remember, the product rule states that, when f(x)=g(x)h(x), \[f'(x)=g'(x)h(x)+g(x)h'(x)\].

Do you get this part?

Answer 8

yes yes, I got that.

Answer 9

So, you need to first find the derivative of e^x and derivative of sin(2x) separately. What are they?

Answer 10

e^x*2cos(2x)

Answer 11

wait wait wait

Answer 12

take your time :D

Answer 13

d/x e^x is e^x and d/dx sin(2x) is 2cos(2x)

Answer 14

That's right. Now you just need to put these values back into the product rule equation.

Answer 15

Do you want to try that, or do you want me to show it to you?

Answer 16

ill make an attempt :)

Answer 17

Good choice

Answer 18

but why do I use product rule first when xsin(2x) is the power?

Answer 19

i mean why use e^x and not just x

Answer 20

Wait wait wait... is your equation\[{ e }^{ xsin(2x) }\quad \quad \quad or\quad \quad \quad e^xsin(2x)\]

Answer 21

e^x2cos2x+e^xsin2x

Answer 22

the first one

Answer 23

Oh.. sorry, I thought it's the second one. Ok. You use the chain rule first. So, \[\frac { d }{ dx } { e }^{ xsin(2x) }={ e }^{ xsin(2x) }\quad \times \quad \frac { d }{ dx } (xsin(2x))\]

Any problem with this?

Answer 24

okay ill show you what I put and let me know how it looks to you

Answer 25

ok

Answer 26

\[e^{xsin2x} * \sin2x +2xcos2x\]

Answer 27

Make sure you put brackets around sin2x+2xcos2x. Other than that, you've got it.

Answer 28

It was marked wrong on my test lol

Answer 29

Then it's your teachers fault. There's this cool website that you can use to check your derivatives: http://www.derivative-calculator.net/

Answer 30

it was worth 1 point and she added the brackets so I guess she felt it was completely wrong without the brackets

Answer 31

or she took away one point for not adding the brackets and that totaled zero. I think she hates me

Answer 32

Well, yea. You will get a different value without the brackets. Without the brackets, it's saying that the e^xsin2x is multiplying 2xcosx, then adding sin2x. But with the bracket, e^xsin2x is multiplying both terms in the bracket.

Answer 33

i still think she hates me.
can you check this one for me f(x) =\[x^{\sin(x)}\] My answer is \[\frac{ 1 }{ x^{sinx}}lnx*cosx\]

Answer 34

lnx goes in the denom

Answer 35

So you mean \[\frac { 1 }{ x^{ sinx }\ln(x) } \times \cos(x)\]

Answer 36

sorry, the cosx goes in the denom too. so everything is in the denom

Answer 37

wow that site is cool. thankx for passing it along

Answer 38

Sorry I've to go now. But that site should show you the steps. Feel free to ask someone else on this site

Answer 39

Didnt check this, but the derivative should look something like this \[ x^{ \sin(x) }(\frac { sinx }{ x } +cosx+lnx)\]

Answer 40

ty

Answer 41

np, cya.

Answer 42

bye