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Math Help - Chain Rule Derivative Question

  1. #1
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    Chain Rule Derivative Question

    Find the derivative of y = (tan(8x))^2.

    Here is what I did, but apparently it's incorrect:

    y' = \frac {dy}{du} \frac{du}{dx} = [\frac{d}{du}u^2][\frac{d}{dx} tan 8x]

    =2[tan 8x][8][\sec ^2 x]
    =16[tan 8x][\sec ^2 x]
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  2. #2
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    Quote Originally Posted by joatmon View Post
    Find the derivative of y = (tan(8x))^2.

    Here is what I did, but apparently it's incorrect:

    y' = \frac {dy}{du} \frac{du}{dx} = [\frac{d}{du}u^2][\frac{d}{dx} tan 8x]

    =2[tan 8x][8][\sec ^2 x]
    =16[tan 8x][\sec ^2 x]
    It should be \sec^2(8x).
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  3. #3
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    Make \displaystyle u=\tan 8x \implies \frac{du}{dx} = 8\sec^28x
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  4. #4
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    u=8x

    w=tanu

    y=w^2

    \displaystyle\frac{dy}{dx}=\frac{dy}{dw}\frac{dw}{  du}\frac{du}{dx}
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