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

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anonymous
  • anonymous
Yes
zepdrix
  • zepdrix
So what's the derivative of cos x? :)
anonymous
  • anonymous
sin x ?
anonymous
  • anonymous
wait -sinx
zepdrix
  • zepdrix
Yes good c:\[\large y'=\color{royalblue}{\left(e^{4x}\right)'}\cos x+e^{4x}\left(-\sin x\right)\]
zepdrix
  • zepdrix
Do you remember the derivative of \(\huge e^x\) ?
anonymous
  • anonymous
Nope :)
zepdrix
  • zepdrix
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. \[\huge \left(e^{4x}\right)'=e^{4x}\left(4x\right)'\]
zepdrix
  • zepdrix
Hmm so what is the derivative of that exponent. The derivative of \(\large 4x\).... hmmmm
anonymous
  • anonymous
4x
zepdrix
  • zepdrix
The derivative of 4x is 4x? Hmm no that's not going to work :c
anonymous
  • anonymous
4
anonymous
  • anonymous
lol
zepdrix
  • zepdrix
Yessss good :) Giving us, \(\huge e^{4x}(4)\) Which gives us an answer of,\[\huge y'=4e^{4x}\cos x+e^{4x}\left(-\sin x\right)\]
anonymous
  • anonymous
WOW...you just made math fun. lol Thanks
zepdrix
  • zepdrix
lol :3 np
anonymous
  • anonymous
wait...its with respect to x

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