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Thread: 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.


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

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

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


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


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


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

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