Question

Differentiate y = e^(4x) cos x with respect to x

dy/dx = ?

Answer 1

Hmm looks like we'll have to apply the product rule :)

Answer 2

You have a product so use the product rule

Answer 3

\[\large y=e^{4x}\cos x\]\[\large y'=\color{royalblue}{\left(e^{4x}\right)'}\cos x+e^{4x}\color{royalblue}{\left(\cos x\right)'}\]

Understand the setup for Product Rule?
We have to take the derivative of the blue terms.

Answer 4

Yes

Answer 5

So what's the derivative of cos x? :)

Answer 6

sin x ?

Answer 7

wait -sinx

Answer 8

Yes good c:\[\large y'=\color{royalblue}{\left(e^{4x}\right)'}\cos x+e^{4x}\left(-\sin x\right)\]

Answer 9

Do you remember the derivative of \(\huge e^x\) ?

Answer 10

Nope :)

Answer 11

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.

\[\huge \left(e^{4x}\right)'=e^{4x}\left(4x\right)'\]

Answer 12

Hmm so what is the derivative of that exponent. The derivative of \(\large 4x\).... hmmmm

Answer 13

4x

Answer 14

The derivative of 4x is 4x? Hmm no that's not going to work :c

Answer 15

4

Answer 16

lol

Answer 17

Yessss good :) Giving us, \(\huge e^{4x}(4)\)





Which gives us an answer of,\[\huge y'=4e^{4x}\cos x+e^{4x}\left(-\sin x\right)\]

Answer 18

WOW...you just made math fun. lol Thanks

Answer 19

lol :3 np

Answer 20

wait...its with respect to x