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Math Help - differentiate tan^3[2x]

  1. #1
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    differentiate tan^3[2x]

    Differentiate tan^32x with respect to x
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  2. #2
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    y = \tan^3{2x}.


    Let u = \tan{2x} so that y = u^3.


    \frac{du}{dx} = 2\sec^2{2x}.


    \frac{dy}{du} = 3u^2 = 3\tan^2{2x}.


    Therefore \frac{dy}{dx} = 6\sec^2{2x}\tan^2{2x}.


    You could use the Pythagorean Identity to clean this up further if you wanted...
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