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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

Mathematics
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Chain rule:

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Other answers:

F(g(x))=F'(g(x))*G'(x)
So the question is: is my statement in the first post correct , if no why not?
Now, you have three functions going on there. Do you see that part?
Honestly I have no idea what's going on , the most difficult thing I ever had to differentiate
Lets rewrite it.\[\tan^3(\sqrt{x})=[tan(x^{\frac{1}{2}})]^3\]Does that help you see three functions?
hold on
that last part is just product rule, so not that hard to do once you know the derivative of this part.
uhm I dunno how you define a function, I cant count it
tan is a function, x with this power is a function, the third?
Any time x has something done to it, it can be seen as a new function.
The whole thing to the 3rd power is the 3rd.
alright and then?
So, you ever seen the chain rule applied multiple times?
just twice
now it needs three times?
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.
that gets me (x^(-1/2)tan^3(sqrt(x)))/2 + sqrt(x)3(tan^2(sqrt(x))2tan(sqrt(x)) ?
\(d/dx~ h(f(n))=h'(f(n))f'(n)\) 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: \(d/dx~ h(f(g(x)))=h'(f(g(x)))f'(g(x))g'(x)\)
\[\large y'=\color{orangered}{(\frac{1}{2}x^{-1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{royalblue}{(\tan^3\sqrt{x})'}\]Hmm, your first derivative looks correct, let's see if we can figure out what's going on with the other one.
ok
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.
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
oh my bad hehe c:
np. It all works. Not like we are contradicting each other. LOL. I have had that happen.
I see how it goes now, lets see what I can do about this
go team go! (Well, a team of 1....)
(x(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(-1/2))/2 oh boy
the second "(" count it as a "^"
(x^(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(-1/2))/2
(x^(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x^(-1/2))/2
\[\large y'=\color{orangered}{(\frac{1}{2}x^{-1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{orangered}{(\large 3\tan^2\sqrt x)(\sec^2\sqrt x)(\frac{1}{2}x^{-1/2})}\] yah it looks like you did it correctly. :) should probably be simplified down a smidge though.
ty bro

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