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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
 10 months ago
 10 months 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
 10 months ago
 10 months ago

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ChristosBest ResponseYou've already chosen the best response.0
@dan815 @e.mccormick @Emily778 @Euler271 @Hero @jim_thompson5910 @Luigi0210 @modphysnoob @primeralph
 10 months ago

Luigi0210Best ResponseYou've already chosen the best response.0
F(g(x))=F'(g(x))*G'(x)
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
So the question is: is my statement in the first post correct , if no why not?
 10 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
Now, you have three functions going on there. Do you see that part?
 10 months ago

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

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

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

ChristosBest ResponseYou've already chosen the best response.0
uhm I dunno how you define a function, I cant count it
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
tan is a function, x with this power is a function, the third?
 10 months ago

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

e.mccormickBest ResponseYou've already chosen the best response.0
The whole thing to the 3rd power is the 3rd.
 10 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
So, you ever seen the chain rule applied multiple times?
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
now it needs three times?
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.1
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.
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
that gets me (x^(1/2)tan^3(sqrt(x)))/2 + sqrt(x)3(tan^2(sqrt(x))2tan(sqrt(x)) ?
 10 months ago

e.mccormickBest 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: \(d/dx~ h(f(g(x)))=h'(f(g(x)))f'(g(x))g'(x)\)
 10 months ago

zepdrixBest 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.
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.1
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.
 10 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
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
 10 months ago

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

ChristosBest ResponseYou've already chosen the best response.0
I see how it goes now, lets see what I can do about this
 10 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
go team go! (Well, a team of 1....)
 10 months ago

ChristosBest 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
 10 months ago

ChristosBest ResponseYou've already chosen the best response.0
the second "(" count it as a "^"
 10 months ago

ChristosBest 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
 10 months ago

ChristosBest 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
 10 months ago

zepdrixBest 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.
 10 months ago
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