Question

What is the derivative of e^(sin2x)? Is it 2cos(2x)e^(sin2x)?

Answer 1

Yes c:
Oh closed... figure it out already?

Answer 2

Actually, no, quick question! Where did the 2 in front of the cos come from? @zepdrix

Answer 3

\[\Large\rm \color{royalblue}{\left[e^{\sin(2x)}\right]'}\]So we have to apply the chain rule a bunch of times here :d

Answer 4

\[\Large\rm \color{royalblue}{\left[e^{\sin(2x)}\right]'}=e^{\sin(2x)}\color{royalblue}{\left[\sin(2x)\right]'}\]So there is our first chain being applied,
we need the derivative of this inner function sin(2x).

Answer 5

\[\Large\rm \color{royalblue}{\left[e^{\sin(2x)}\right]'}=e^{\sin(2x)}\cos(2x)\color{royalblue}{\left[2x\right]'}\]Then chain one last time :)

Answer 6

See where that pesky 2 is coming from?

Answer 7

Hopefully the coloring made sense.
Blue represents something we need to differentiate*

Answer 8

So is the [2x]' from the (2x) in the exponent?

Answer 9

I guess it's more accurate to say that it's coming from the sin(2x).
The (2x) inside of the sine function let's us know we have to chain rule again.

Answer 10

Got it! Thanks! :)

Answer 11

It's coming from this guy: \(\Large\rm \color{royalblue}{\left[\sin(2x)\right]'}\)
After differentiating the outer function (sine), we have to chain.

Answer 12

Chain rule is really freaky and weird!
Once you get a good handle on it though, you'll love it.
It'll be your best friend.

Answer 13

I'll try to make amends with it. :)