hello guys im kind of confused and my book those not give me answer to this problem i wonder if someone can confirm my answer or if I'm simply wrong. If I'm wrong then please correct me, don't just tell me I'm wrong. I'm asked to find the derivative of the following
$$ y= e^{-2t} \cos4t$$
here is how i did it
step 1 by the product rule-
$$y'= e^{-2t} \frac{d}{dt}(\cos(4t))+\cos(4t) \frac{d}{dt} (e^{-2t}) $$
step 2 – find the derivatives of the two functions using chain rule
$$e^{-2t}(\cos(4t)(-\sin(4t)(4)))+ \cos(4t)(e^{-2t}(-2))$$
did i approached this correctly? is this the correct answer.
Thanks in advance of any adivce you guys can offer. Im really leary about the $(\cos(4t)(-\sin(4t)(4)))$ im not sure if this part is right or not?
Thanks
Miguel
Best Answer
It should be $$ e^{-2t}(-\sin(4t) \cdot 4)+ \cos(4t)(e^{-2t}(-2)).$$ Your logic is right, but $$\frac{d}{dt}(\cos(4t)) = -4\sin(4t),$$ not $-4\sin(4t)\cos(4t).$