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Math Help - Use implicit differentiation on x=cot(y) to find dy/dx in terms of x

  1. #1
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    Use implicit differentiation on x=cot(y) to find dy/dx in terms of x

    I have x'=-csc^2y

    Or dy/dx=-csc^2y

    Am I doing this problem right?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    you are differntiating wrt x

    d(x)/dx= d[cot(y)]/dx

    1= -csc^2(y) dy/dx

    dy/dx= -1/csc^2(y) = -sin^2(y)


    Alternatively y = arcot(x)

    dy/dx = -1/ (1+x^2)


    Since y= arcot(x) draw a right triangle with an acute angle t whose adjacent side is x and whose opposite side is 1 making the hypoteneuese sqrt(1+x^2) so sin^2(y) = 1 /(1+x^2)

    so our previous result dy/dx = -sin^2(y) = -1/ (1+x^2) is consistent with our alternative approach
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