OpenStudy (anonymous):

Differentiate e^xsin(2x)

4 years ago
OpenStudy (anonymous):

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)

4 years ago
OpenStudy (anonymous):

could you please show me.

4 years ago
OpenStudy (anonymous):

Do you know the product rule?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

Do you mind showing me your working?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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?

4 years ago
OpenStudy (anonymous):

yes yes, I got that.

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

e^x*2cos(2x)

4 years ago
OpenStudy (anonymous):

wait wait wait

4 years ago
OpenStudy (anonymous):

take your time :D

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

ill make an attempt :)

4 years ago
OpenStudy (anonymous):

Good choice

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

i mean why use e^x and not just x

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

e^x2cos2x+e^xsin2x

4 years ago
OpenStudy (anonymous):

the first one

4 years ago
OpenStudy (anonymous):

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?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

ok

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

It was marked wrong on my test lol

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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.

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

lnx goes in the denom

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

wow that site is cool. thankx for passing it along

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

ty

4 years ago
OpenStudy (anonymous):

np, cya.

4 years ago
OpenStudy (anonymous):

bye

4 years ago