## skay 2 years ago can someone help me with finding the derivative of: f(x)=(e^-2t)cos(4t) ?

use the product rule, also the chain rule if given y = uv then y ' = u'v + uv' let u=e^(-2t) ----> u' = ... ? v =cos4t ----> v' = ... ?

2. skay

thank you :) $4\sin (4t)$ is this the derivative of cos(4t)?

3. zepdrix

$$\large \cos(4t) \qquad \rightarrow \qquad -4\sin(4t)$$ You were close! :) Just missed the negative when cosine goes to sine.

4. skay

yay thanks. :) I'm having trouble with the chain rule. I got to the derivatives of both sides but im not sure what to do with $-2te ^{-2t-1}(e ^{-2t}\cos(4t))-4\sin(4t)$

5. zepdrix

Woops! Little boo boo on the first part. You can not apply the power rule to an exponential function.

6. skay

well crud. what do i do with $e ^{-2t}$ I honestly dont remember.

7. zepdrix

Remember this derivative? :) $$\large (e^t)'=e^t$$ We'll be doing the same thing, but since our exponent is more than just t, we need to apply the chain rule, multiply by the derivative of the exponent. Example,$\large (e^{3t})' \qquad = \qquad e^{3t}(3t)' \qquad = \qquad e^{3t}(3)$

8. skay

$-2e ^{-2t}(e ^{-2t}\cos(4t))-4\sin(4t)$ Hmmm. I hope this is what you're talking about. I feel like the intial equation in the middle should be comething different :(

9. zepdrix

$\large f(t)=(e^{-2t})\cos4t$ $\large f'(t)=\color{royalblue}{(e^{-2t})'}\cos4t+(e^{-2t})\color{royalblue}{(\cos4t)'}$Here is the setup for the product rule. We need to differentiate the blue terms.$\large f'(t)=\color{orangered}{(-2e^{-2t})}\cos4t+(e^{-2t})\color{royalblue}{(\cos4t)'}$That's the derivative of the first blue term.$\large f'(t)=\color{orangered}{(-2e^{-2t})}\cos4t+(e^{-2t})\color{orangered}{(-4\sin4t)}$There is the other one.

10. zepdrix

I can't quite tell what is going on with yours. Not sure why there are 2 exponentials on the first term.. hmm

11. skay

thank you so much :) i didnt quite understand the rules. I think ive solved it. $-2 e ^{-2t} (\cos(4 t)+2 \sin(4 t))$

12. zepdrix

cool c: