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Math Help - partial differentiation (trig)

  1. #1
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    partial differentiation (trig)


    Im trying to find the partial derivative of df/dy, but im not sure how to differentiate tan^-1 (x/y).
    could someone check if i differentiated the log correctly?
    Thanks!
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by gomes View Post

    Im trying to find the partial derivative of df/dy, but im not sure how to differentiate tan^-1 (x/y).
    could someone check if i differentiated the log correctly?
    Thanks!
    Yes your differentiation is correct.

    Let  p = tan^{-1} \frac{x}{y}

     tanp = \frac{x}{y}

    Implicitly differentiate

     (sec^2 p) p^{ \prime} = - \frac{x}{y^2}

     p^{ \prime} = - \frac{x}{y^2 sec^2 p }

     p^{ \prime} = - \frac{x}{y^2 sec^2 (tan^{-1} \frac{x}{y} ) }
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  3. #3
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    thanx
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  4. #4
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    That's correct but a very strange way of writing the answer!

    I would, instead, have used the fact that the derivative of tan^{-1}(x) is \frac{1}{x^2+ 1}. Let u= \frac{x}{y} so that \frac{\partial u}{\partial y}= \frac{\partial \frac{x}{y}}{\partial y}= -\frac{x}{y^2}. Also tan^{-1}(\frac{x}{y})= tan^{-1}(u) and \frac{\partial tan^{-1}(u)}{\partial u}= \frac{1}{u^2+ 1}.

    By the chain rule, then, \frac{\partial tan^{-1}(\frac{x}{y})}{\partial y} = \frac{1}{u^2+ 1}\left(-\frac{x}{y^2}\right)=  -\frac{1}{\frac{x^2}{y^2}+ 1}\frac{x}{y^2}= -\frac{x}{x^2+ y^2}.

    But, as I said, AllenCuz's result is also correct- These answers are exactly the same.

    Imagine a right triangle with angle \theta, "opposite side" of length x, and "near side" of length y. Then tan(\theta)= \frac{x}{y} so [tex]/theta= tan^{-1}(\frac{x}{y}). But then the hypotenuse of the triangle has length \sqrt{x^2+ y^2} and sec(\theta)= sec(tan^{-1}(\frac{y}{x}))= \frac{\sqrt{x^2+ y^2}}{y} so that sec^2(tan^{-1}(\frac{y}{x}))= \frac{x^2+ y^2}{y^2}.

    -\frac{x}{y^2 sec(tan^{-2}(\frac{y}{x}))}= -\frac{x}{y^2}\frac{y^2}{x^2+ y^2}= -\frac{x}{x^2+ y^2}.
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