Differentiate y = e^(4x) cos x with respect to x dy/dx = ?

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Differentiate y = e^(4x) cos x with respect to x dy/dx = ?

Calculus1
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Hmm looks like we'll have to apply the product rule :)
You have a product so use the product rule
\[\large y=e^{4x}\cos x\]\[\large y'=\color{royalblue}{\left(e^{4x}\right)'}\cos x+e^{4x}\color{royalblue}{\left(\cos x\right)'}\] Understand the setup for Product Rule? We have to take the derivative of the blue terms.

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Other answers:

Yes
So what's the derivative of cos x? :)
sin x ?
wait -sinx
Yes good c:\[\large y'=\color{royalblue}{\left(e^{4x}\right)'}\cos x+e^{4x}\left(-\sin x\right)\]
Do you remember the derivative of \(\huge e^x\) ?
Nope :)
Oooo tsk tsk! That's a fun easy one that you'll want to remember! c: \[\huge \left(e^x\right)'=e^x\] It gives us the same thing back. This same thing will happen in our problem here, except we'll have an extra step. Since our exponent is more than just \(\large x\), we have to apply the chain rule. Multiply the result by the derivative of the exponent. \[\huge \left(e^{4x}\right)'=e^{4x}\left(4x\right)'\]
Hmm so what is the derivative of that exponent. The derivative of \(\large 4x\).... hmmmm
4x
The derivative of 4x is 4x? Hmm no that's not going to work :c
4
lol
Yessss good :) Giving us, \(\huge e^{4x}(4)\) Which gives us an answer of,\[\huge y'=4e^{4x}\cos x+e^{4x}\left(-\sin x\right)\]
WOW...you just made math fun. lol Thanks
lol :3 np
wait...its with respect to x

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