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Math Help - Implicit Differentiation of Trig functions

  1. #1
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    Implicit Differentiation of Trig functions

    How would one go about doing these? I know how to do implicit differentiation, just not of trig functions. Specifically I'm trying to find y sec(x) = x tan(y)
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  2. #2
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    Hello, fattydq!

    I must assume you know the trig derivatives . . . and the Product Rule.


    y\cdot\sec(x) \:=\:x\cdot\tan(y)

    . . . . . . \underbrace{y\cdot\sec(x)}_{\text{Product Rule}} \qquad\qquad = \qquad\quad\;\; \underbrace{x\cdot\tan(y)}_{\text{Product rule}}

    \overbrace{y\cdot\sec(x)\tan(x) +  y'\cdot\sec(x)} \;=\; \overbrace{x\cdot\sec^2(y)\cdot y' + 1\cdot\tan(y)}


    . . \sec(x)\cdot y' - x\sec^2(y)\cdot y' \;=\;\tan (y) - y\sec(x)\tan(x)


    Factor: . \bigg[\sec(x) - x\sec^2(y)\bigg]\cdot y' \;=\;\tan(y) - y\sec(x)\tan(x)


    Therefore: . y' \;=\;\frac{\tan(y) - y\sec(x)\tan(x)}{\sec(x) - x\sec^2(y)}

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