Question

Please help me!
Differentiate:

Answer 1

\[y=\frac{ \cosh(\cos2x) }{ \arctan \sqrt{x} }\]

Answer 2

Woah! This one is a doozy XD

Answer 3

Hmm doesn't look like we have any nice simplification, so jump right in using the quotient rule.

Answer 4

\[\large y'=\frac{[\cosh (\cos 2x)]'\arctan \sqrt{x} - \cosh (\cos 2x)[\arctan \sqrt x]'}{(\arctan \sqrt x)^2}\]

Answer 5

okay!

Answer 6

So the setup is quite messy... but besides that, we don't have too much work to do. We just need to differentiate the 2 primed terms.

Maybe we should do those off to the side.
The hyperbolic cosine is going to have a few chains to it. Hmm.

Answer 7

Remember the derivative of cosh? :D

Answer 8

yes! sinh

Answer 9

\[\large [\cosh (\cos 2x)]'=[\sinh (\cos 2x)](\cos 2x)'\]
Yah sounds good c: that takes care of the outer function.

Answer 10

okay!

Answer 11

I just don't know how to do the derivative of \[\arctan \sqrt{x}\]

Answer 12

So we're going to stuff it into this form, then apply the chain rule as we usually would.\[\huge (\arctan z)'=\frac{1}{1+z^2}z'\]

Answer 13

\[\huge (\arctan \sqrt x)'=\frac{1}{1+(\sqrt x)^2}(\sqrt x)'\]
Understand what we did there?

Answer 14

yupp! What would the derivative of \[\sqrt{x}\] be?

Answer 15

oh comeon morgan! really? -_- lol

Answer 16

you should have that one memorized by now ^^ hehe

If you can't remember, change it to a fractional exponent :D then apply the power rule.

Answer 17

OH! sorry, i had a brain fart! lol

Answer 18

would the answer be:
\[y \prime=\frac{ 2(\arctan \sqrt{x})(\sinh)(\sin2x)-\cosh(\cos2x)(\frac{ 1 }{ 1+x })(1/2x ^{-1/2}}{ (\arctan \sqrt{x})^{2} }\]

Answer 19

Hmmm we're close.
Your sinh term doesn't have anything written inside of it. :O hmmm

Answer 20

what do you mean?

Answer 21

the sinh function should have an inside.. you ripped his guts out, where did they go? :(

Answer 22

Okay, I get it. My brother just helped me! Lol. Can you help me with the derivative of \[3^{arcsinx}\] ??

Answer 23

Yay, brother to the rescue! :D

Answer 24

Remember how to take the derivative of an exponential with base that IS NOT e? Here is a quick refresher just in case.
\[\huge (3^x)'=3^x(\ln 3)\]

Answer 25

i get it!

Answer 26

Just don't forget to apply the chain rule to the exponent!\[\huge (3^{\arcsin x})'=(3^{\arcsin x})(\ln 3)(\arcsin x)'\]

Answer 27

Woops that was too big lemme try that again.
\[\large (3^{\arcsin x})'=(3^{\arcsin x})(\ln 3)(\arcsin x)'\]