Question

answer for cookies!
f(t) = tan(e^3t) + e^tan 3t

Answer 1

Oooo cookies :OOO

Answer 2

What are we going? Differentiating?

Answer 3

yes sir

Answer 4

Lady bunch! I thought you already did this problem a few days ago! >.<
Maybe I'm mistaken though...

Answer 5

ran out of cookies, had to do it again

Answer 6

\[\Large\rm f(t) = \tan(e^{3t}) + e^{\tan 3t}\]

Answer 7

Do you remember your derivatives for e^x and tan x?

Answer 8

e^x equal itself tanx is sec^2x

Answer 9

Mmm k good good.

Answer 10

Sorry power went out :c Did I lose my cookie.
Ok ok we're still good.. sec

Answer 11

So using your knowledge of those derivatives,
it tells us that when we differentiate we should get something like:
\[\Large\rm f'(t) = \sec^2(e^{3t}) + e^{\tan 3t}\]

But we need to apply our chain rules, yes?

Answer 12

\[\Large\rm f'(t) = \sec^2(e^{3t})\color{royalblue}{\left(e^{3t}\right)'}+ e^{\tan 3t}\color{royalblue}{\left(\tan 3t\right)'}\]So we need to multiply by the derivative of the inner functions.

Answer 13

So again, as you said before, our exponential gives us the same thing back:\[\Large\rm f'(t) = \sec^2(e^{3t})\color{orangered}{\left(e^{3t}\right)}\color{royalblue}{(3t)'}+ e^{\tan 3t}\color{royalblue}{\left(\tan 3t\right)'}\]But we have to chain rule again!! :O

Answer 14

\[\Large\rm f'(t) = \sec^2(e^{3t})\color{orangered}{\left(e^{3t}\right)}\color{orangered}{(3)}+ e^{\tan 3t}\color{royalblue}{\left(\tan 3t\right)'}\]

Answer 15

So we've fully differentiated the first term.
How bout that other term, Baby Bump? What do you think? :)
Give it a try!
What should we get?

Answer 16

Comeon Banana Bunch! Don't give up on me now >:U
Participateeeee

Answer 17

\[\sec^23t(3)\]

Answer 18

Yay good job \c:/