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Christos
 one year ago
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
 one year ago
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

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Christos
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 @e.mccormick @Emily778 @Euler271 @Hero @jim_thompson5910 @Luigi0210 @modphysnoob @primeralph

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0F(g(x))=F'(g(x))*G'(x)

Christos
 one year ago
Best ResponseYou've already chosen the best response.0So the question is: is my statement in the first post correct , if no why not?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Now, you have three functions going on there. Do you see that part?

Christos
 one year ago
Best ResponseYou've already chosen the best response.0Honestly I have no idea what's going on , the most difficult thing I ever had to differentiate

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Lets rewrite it.\[\tan^3(\sqrt{x})=[tan(x^{\frac{1}{2}})]^3\]Does that help you see three functions?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0that last part is just product rule, so not that hard to do once you know the derivative of this part.

Christos
 one year ago
Best ResponseYou've already chosen the best response.0uhm I dunno how you define a function, I cant count it

Christos
 one year ago
Best ResponseYou've already chosen the best response.0tan is a function, x with this power is a function, the third?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Any time x has something done to it, it can be seen as a new function.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0The whole thing to the 3rd power is the 3rd.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0So, you ever seen the chain rule applied multiple times?

Christos
 one year ago
Best ResponseYou've already chosen the best response.0now it needs three times?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.0that gets me (x^(1/2)tan^3(sqrt(x)))/2 + sqrt(x)3(tan^2(sqrt(x))2tan(sqrt(x)) ?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0\(d/dx~ h(f(n))=h'(f(n))f'(n)\) 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\large y'=\color{orangered}{(\frac{1}{2}x^{1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{royalblue}{(\tan^3\sqrt{x})'}\]Hmm, your first derivative looks correct, let's see if we can figure out what's going on with the other one.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let'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
 one year ago
Best ResponseYou've already chosen the best response.0A 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

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0np. It all works. Not like we are contradicting each other. LOL. I have had that happen.

Christos
 one year ago
Best ResponseYou've already chosen the best response.0I see how it goes now, lets see what I can do about this

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0go team go! (Well, a team of 1....)

Christos
 one year ago
Best ResponseYou've already chosen the best response.0(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
 one year ago
Best ResponseYou've already chosen the best response.0the second "(" count it as a "^"

Christos
 one year ago
Best ResponseYou've already chosen the best response.0(x^(1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(1/2))/2

Christos
 one year ago
Best ResponseYou've already chosen the best response.0(x^(1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x^(1/2))/2

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large y'=\color{orangered}{(\frac{1}{2}x^{1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{orangered}{(\large 3\tan^2\sqrt x)(\sec^2\sqrt x)(\frac{1}{2}x^{1/2})}\] yah it looks like you did it correctly. :) should probably be simplified down a smidge though.
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