Question

(Calc Q > Derivatives > Chain Rule > Double chain rule scenarios?!)
So I'm working on some problems for my homework and come across a problem that requires me to apply the chain rule twice. I have no clue where to begin, even though I do understand how to apply the chain rule. The problem in question is:
g(x) = 2tan^3(4πx).
The answer is 24π(tan^2(4πx))(sec^2(4πx).

However having the answer helps me none since I do not know how to apply the chain rule twice... Help?! (MUCH thanks in advance!!!)

Answer 1

Here's a step by step process.\[g(x) = 2\tan^3(4πx)\]\[{d \over dx}\quad 2\tan^3(4πx) \]\[=6\tan^2(4πx) \cdot {d \over dx}\;\;\tan(4\pi x)\]\[=6\tan^2(4πx) \cdot \sec^2(4 \pi x) \cdot {d \over dx}\;\;4\pi x\]\[=6\tan^2(4πx) \cdot \sec^2(4 \pi x) \cdot 4\pi\]\[=24 \pi \cdot \tan^2(4πx) \cdot \sec^2(4 \pi x)\]

Answer 2

f(g(h(x)))'
= f'(g(h(x))*g(h(x))'
g(h(x))' = g'(h(x))*h'(x)
So I get:
f'(g(h(x))*g'(h(x))*h'(x)

Answer 3

You basically just slowly move the \[d \over dx\]through the entire equation integrating with the chain rule one part at a time.

Answer 4

Basically, just start on the outside function, take the derivative, and plug all of the inside functions into that derivative. Then multiply the result by the derivative of everything inside.

Answer 5

erret, lets work on that... and while im trying this, could you come up with something else for me to try so i know that I get this?

Answer 6

Yes.
cos(sin(x^2)

Answer 7

tan(4πx) comes from where? second part third step

Would that be the inside of the original equasion?

Answer 8

like what I don't understand is that usually, you would take d/dx(outside) x (inside) x d/dx(inside), as in d/dx(2(x^2-1)^2 ) = 4(x^2-1)(2x)

Answer 9

Rewrite the function as:
2tan(4pix)^3
Chain rule says I can take the derivative as if it was just 2x^3, as long as I multiply by the derivative of the inside function after I do that. The inside in this case is tan(4pix)

Answer 10

The first function you were integrating is \(u^3\) where \(u=\tan(4\pi x)\) So we take the derivative to get \[{d \over du} u^3=3u^2 \cdot \left({d \over du}\;\; u\right)\]By the definition of the chain rule. Now substitute \(u=\tan(4\pi x)\) back in, to get the expression that I wrote.

Answer 11

This makes much more sense, now I shall attempt smoothmath's problem... btw thanks guys.

Answer 12

You're welcome.

Answer 13

My pleasure =D

Answer 14

alright so the problem, while somewhat similar to what ye gents described... well lol here is what I got?

-sin(sinx^2)(cos(x^2)(2x))

I'm double-checking my work as I post, just wanted to get this out there

Answer 15

That looks correct to me.

Answer 16

spit out one more with slightly more complex numbers? :)

Answer 17

How about this? \[-\cos(5\sin^2(x^4)+\tan(x^2))\]

Answer 18

:D one moment please!

Answer 19

I also agree with your answer for that first one.

Answer 20

i'm going to try to apply the method used in http://www.youtube.com/watch?v=8qx5XsyXBQA , so i may take a little longer

Answer 21

People who read cos as "cose" instead of "cosine"
cringe...

Answer 22

sin(5sin(x^8+tanx^2)(40sinx^7(cosx)+2(tanx)(sec^2x) is basically it, but i got careless with my parenthesis so i need to re-establish where the **** they need to be added to close the equasion -.-

Answer 23

It looks close, but there's a few terms that are off. I certainly gave you a hard one.

Answer 24

that would be agreeable. wolfram alpha also think that i am wrong, but I am a little lost on a few features on this answer..

>> http://www.wolframalpha.com/input/?i=d%2Fdx [%E2%88%92cos%285sin2%28x4%29%2Btan%28x2%29%29] <<

first things first, can 5sin^2(x^4) simplify to 5sin(x^4)^2)? and then must I double chain rule this or can I multiply in the ^2 to the ^4 to get 5sinx^8?

Answer 25

You have to use the double chain rule on that.

Answer 26

blehhhhhhhhhhhhhhhhhhhhhhhh lol okay. Well... thanks a ton for your help!! I think I am alright from here. :D

Answer 27

You're very welcome. If you want a slightly easier problem to work, try \[-\cos(5\sin^2(x^4))\]Instead