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Mathematics 16 Online
OpenStudy (anonymous):

find f' for the function: f(x)= e^2xcos3x

OpenStudy (anonymous):

Not clear what the function is. Is it\[f(x)= e^{2xcos3x}~or~f(x)= e^{2x}\cos3x\]

OpenStudy (anonymous):

the second one. i have to use the rule (f x g)'= f' x g + f x g'

OpenStudy (anonymous):

Correct. What do you get?

OpenStudy (anonymous):

Make sure you use the chain rule along the way....

OpenStudy (anonymous):

i have e^2x = 2e^2x cos 3x but i forgot how to find (cos 3x)' i know it becomes -sin but i dont know how to combine that with e^2x

OpenStudy (anonymous):

I get \[f(x)=e^{2x}\cos3x \implies f'(x)=2e^{2x}\cos3x−3e^{2x}\sin3x\]

OpenStudy (anonymous):

the only thing i cant remember is why there is a 3 in front of e^2x sin3x

OpenStudy (anonymous):

Chain rule I mentioned above. Function is cos(u), derivative is -sin(u) u'. In this case u'=3.

OpenStudy (anonymous):

Came from the same place the two in front of the first term did.

OpenStudy (anonymous):

i get it now. now to find f' of the new function, ill just use the chain rule again?

OpenStudy (anonymous):

f=2e^2x cos3x g=-3e^2x sin3x

OpenStudy (anonymous):

Stated another way, let f(x) = g(x)h(x), so f'(x)= g'h(x)+gh'(x). In the current case, g(x)=e^(2x) and h(x)=cos(3x). Use the chain rule on g and h to get g'(x)=2e^(2x) and h'(x)=-3sin(3x). Assemble as indicated above to get the answer I have further above.

OpenStudy (anonymous):

You want to find the second derivative?

OpenStudy (anonymous):

unfortunately :-(

OpenStudy (anonymous):

No sweat, just keep doing it again. Remember, you have two terms in the first derivative, each with two functions, so your second derivative will have four terms before simplification.

OpenStudy (anonymous):

i like the way you explained it before using "h" and "g". its more clear somehow

OpenStudy (anonymous):

i got: 2 [2e^(2x) cos3x-3e^(2x) sin3x] - 3[2e^(2x) sin3x + 3e^(2x) cos3x] four terms? is that right?

OpenStudy (anonymous):

Lets look. I get this.\[f′(x)=2e^{2x}\cos(3x)−3e^{2x}\sin(3x) \]\[\implies f"(x)=4e^{2x}\cos(3x)-6e^{2x}\sin(3x)-6e^{2x}\sin(3x)-9e^{2x}\cos(3x)\]\[=4e^{2x}\cos(3x)-12e^{2x}\sin(3x)-9e^{2x}\cos(3x)\]

OpenStudy (anonymous):

I think our answers are equivalent.

OpenStudy (anonymous):

These problems aren't too bad as long as you stay methodical and work them step-by-step. Pay attention to the details, and you will get a lot of right answers.

OpenStudy (anonymous):

OMG!!! thank you so much, i feel a weight lifted lol. i was so nervous typing my answer. i want to thank you for your help, i really appreciate it. i have to go now, my moms yelling its dinner time. Thanks again!

OpenStudy (anonymous):

No sweat.

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