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Christos

  • 2 years 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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  1. Christos
    • 2 years ago
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    @zepdrix

  2. Christos
    • 2 years ago
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    @dan815 @e.mccormick @Emily778 @Euler271 @Hero @jim_thompson5910 @Luigi0210 @modphysnoob @primeralph

  3. Luigi0210
    • 2 years ago
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    Chain rule:

  4. Luigi0210
    • 2 years ago
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    F(g(x))=F'(g(x))*G'(x)

  5. Christos
    • 2 years ago
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    So the question is: is my statement in the first post correct , if no why not?

  6. e.mccormick
    • 2 years ago
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    Now, you have three functions going on there. Do you see that part?

  7. Christos
    • 2 years ago
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    Honestly I have no idea what's going on , the most difficult thing I ever had to differentiate

  8. Christos
    • 2 years ago
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    @zepdrix

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

  10. Christos
    • 2 years ago
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    hold on

  11. e.mccormick
    • 2 years ago
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    that last part is just product rule, so not that hard to do once you know the derivative of this part.

  12. Christos
    • 2 years ago
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    uhm I dunno how you define a function, I cant count it

  13. Christos
    • 2 years ago
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    tan is a function, x with this power is a function, the third?

  14. e.mccormick
    • 2 years ago
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    Any time x has something done to it, it can be seen as a new function.

  15. e.mccormick
    • 2 years ago
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    The whole thing to the 3rd power is the 3rd.

  16. Christos
    • 2 years ago
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    alright and then?

  17. e.mccormick
    • 2 years ago
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    So, you ever seen the chain rule applied multiple times?

  18. Christos
    • 2 years ago
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    just twice

  19. Christos
    • 2 years ago
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    now it needs three times?

  20. zepdrix
    • 2 years ago
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    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.

  21. Christos
    • 2 years ago
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    that gets me (x^(-1/2)tan^3(sqrt(x)))/2 + sqrt(x)3(tan^2(sqrt(x))2tan(sqrt(x)) ?

  22. e.mccormick
    • 2 years ago
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    \(d/dx~ h(f(n))=h'(f(n))f'(n)\) 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: \(d/dx~ h(f(g(x)))=h'(f(g(x)))f'(g(x))g'(x)\)

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

  24. Christos
    • 2 years ago
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    ok

  25. zepdrix
    • 2 years ago
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    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.

  26. e.mccormick
    • 2 years ago
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    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

  27. zepdrix
    • 2 years ago
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    oh my bad hehe c:

  28. e.mccormick
    • 2 years ago
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    np. It all works. Not like we are contradicting each other. LOL. I have had that happen.

  29. Christos
    • 2 years ago
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    I see how it goes now, lets see what I can do about this

  30. e.mccormick
    • 2 years ago
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    go team go! (Well, a team of 1....)

  31. Christos
    • 2 years ago
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    (x(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(-1/2))/2 oh boy

  32. Christos
    • 2 years ago
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    the second "(" count it as a "^"

  33. Christos
    • 2 years ago
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    (x^(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(-1/2))/2

  34. Christos
    • 2 years ago
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    (x^(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x^(-1/2))/2

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

  36. Christos
    • 2 years ago
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    ty bro

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