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Math Help - trig derivative

  1. #1
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    trig derivative

    Am I doing this one correct?

    find dy/dx of \tan (xy) = y

    \frac{d}{dx} [y = \tan (xy)]

    1\frac{dy}{dx} = \sec^2 (xy)[y+x\frac{dy}{dx}]

    -x \sec^2 (xy) \frac{dy}{dx} + 1 \frac{dy}{dx} = y\sec^2 (xy)

    \frac{dy}{dx} = \frac{y\sec^2 (xy)}{1-x\sec^2 (xy)}
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  2. #2
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    Re: trig derivative

    Yes, that is correct.
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