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Math Help - Trigonometric differentiation help!

  1. #1
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    Trigonometric differentiation help!

    Hi, I'm having a lot of trouble solving the following question related to trig differentiation. I would greatly appreciate any help.

    1) Differentiate tan^2 (cos x)

    Thanks



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  2. #2
    Senior Member nikhil's Avatar
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    -2(sinx)tan(cosx)[sec(cosx)]^2
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  3. #3
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    Quote Originally Posted by nikhil View Post
    -2(sinx)tan(cosx)[sec(cosx)]^2
    Yes, I'm pretty sure thats the answer..Would you mind sharing a bit of the working?
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  4. #4
    Senior Member nikhil's Avatar
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    d/dx (tan(cosx))^2
    = 2tan(cosx)[d/dx(tan(cosx))]
    =2tan(cosx)(sec(cosx))^2(d/dx (cosx))
    =2tan(cosx)(sec(cosx))^2(-sinx)
    =-2(sinx)tan(cosx)(sec(cosx))^2
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  5. #5
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    Nikhil used the chain rule.

    Let u= cos(x),[/tex]v= tan(u)[/tex], and w= v^2

    Then \frac{dw}{dv}= 2v, \frac{dv}{du}= sec^2(u), and \frac{du}{dx}= - sin(x)

    The chain rule says
    \frac{dw}{dx}= \frac{dw}{dv}\frac{dv}{du}\frac{du}{dx}
    = (2v)(sec^2(u))(-sin(x))= (2tan(u))(sec^2(cos(x))(-sin(x)) = -2tan(cos(x))sec^2(cos(x))sin(x).

    Of course, nikhil did it with specifically writing out "u", "v", and "w".
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