the derivative of tan^3(sqrt(x))
is 3tan^2(sqrt(x))2tan(sqrt(x))
?
I am asking in an attempt to actually find this derivative http://screencast.com/t/1uOEBG1WH

- Christos

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- Christos

@zepdrix

- Christos

@dan815 @e.mccormick @Emily778 @Euler271 @Hero @jim_thompson5910 @Luigi0210 @modphysnoob @primeralph

- Luigi0210

Chain rule:

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## More answers

- Luigi0210

F(g(x))=F'(g(x))*G'(x)

- Christos

So the question is: is my statement in the first post correct , if no why not?

- e.mccormick

Now, you have three functions going on there. Do you see that part?

- Christos

Honestly I have no idea what's going on , the most difficult thing I ever had to differentiate

- Christos

@zepdrix

- e.mccormick

Lets rewrite it.\[\tan^3(\sqrt{x})=[tan(x^{\frac{1}{2}})]^3\]Does that help you see three functions?

- Christos

hold on

- e.mccormick

that last part is just product rule, so not that hard to do once you know the derivative of this part.

- Christos

uhm I dunno how you define a function, I cant count it

- Christos

tan is a function, x with this power is a function, the third?

- e.mccormick

Any time x has something done to it, it can be seen as a new function.

- e.mccormick

The whole thing to the 3rd power is the 3rd.

- Christos

alright and then?

- e.mccormick

So, you ever seen the chain rule applied multiple times?

- Christos

just twice

- Christos

now it needs three times?

- zepdrix

The setup will be the same as the last problem ms christos :o
\[\large y=\sqrt{x}\tan^3\sqrt{x}\]
\[\large y'=\color{royalblue}{(\sqrt{x})'}\tan^3\sqrt{x}+\sqrt{x}\color{royalblue}{(\tan^3\sqrt{x})'}\]
Remember how we did that? Product rule setup.

- Christos

that gets me (x^(-1/2)tan^3(sqrt(x)))/2 + sqrt(x)3(tan^2(sqrt(x))2tan(sqrt(x))
?

- e.mccormick

\(d/dx~ h(f(n))=h'(f(n))f'(n)\)
But what if n=g(x)? Then f'(n) is another chain rule!
\(d/dx~ f(g(x))=f'(g(x))g'(x)\)
When I put the two together:
\(d/dx~ h(f(g(x)))=h'(f(g(x)))f'(g(x))g'(x)\)

- zepdrix

\[\large y'=\color{orangered}{(\frac{1}{2}x^{-1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{royalblue}{(\tan^3\sqrt{x})'}\]Hmm, your first derivative looks correct, let's see if we can figure out what's going on with the other one.

- Christos

ok

- zepdrix

Let's just look at the blue term for now,\[\large \color{royalblue}{(\tan^3\sqrt x)'}\]
So the outermost function is (something) cubed. Applying the power rule and chain rule gives us,\[\large 3(\tan^2\sqrt x)\color{royalblue}{(\tan\sqrt x)'}\]Three came down, giving us a 2 exponent, then we apply the chain rule.
Remember the derivative of tangent?\[\large 3(\tan^2\sqrt x)(\sec^2\sqrt x)\color{royalblue}{(\sqrt x)'}\]
Are you following my notation ok?
The blue terms are ones that we need to take a derivative of.
They keep showing up due to the chain rule.

- e.mccormick

A lot of people like to start at the inner most function, and work their way out. It is not a bad choice because it works well for multiple chain rules. Or work from outer in. However, as zepdrix points out, there is a problem with part of what you did for a derivative. What is the derivative of tangent?
hmm.. he got it as I was typing in my word processor. LOL

- zepdrix

oh my bad hehe c:

- e.mccormick

np. It all works. Not like we are contradicting each other. LOL. I have had that happen.

- Christos

I see how it goes now, lets see what I can do about this

- e.mccormick

go team go! (Well, a team of 1....)

- Christos

(x(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(-1/2))/2
oh boy

- Christos

the second "(" count it as a "^"

- Christos

(x^(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(-1/2))/2

- Christos

(x^(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x^(-1/2))/2

- zepdrix

\[\large y'=\color{orangered}{(\frac{1}{2}x^{-1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{orangered}{(\large 3\tan^2\sqrt x)(\sec^2\sqrt x)(\frac{1}{2}x^{-1/2})}\]
yah it looks like you did it correctly. :) should probably be simplified down a smidge though.

- Christos

ty bro

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