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Maths Differentiation Problem

  • 28-09-2014 5:49pm
    #1
    Registered Users Posts: 23


    Find dy/dx if
    y= sin^-1(cosx)

    It says the answer is -1.
    I've gotten as far as

    cos2(2x) / cos2(2x) - cos2 (2x) X -sinx but not sure where to go from there, or if that's right.


Comments

  • Registered Users, Registered Users 2 Posts: 13,031 ✭✭✭✭bnt


    If it's any help, when you differentiate sin^-1 (x), you get 1 / √(1-x²) a.k.a. (1-x²)^(-½). One of those Calculus Rules may be needed at this point.

    Death has this much to be said for it:
    You don’t have to get out of bed for it.
    Wherever you happen to be
    They bring it to you—free.

    — Kingsley Amis



  • Registered Users, Registered Users 2 Posts: 5,629 ✭✭✭TheBody


    If [latex]y=sin^{-1}(cos(x))[/latex],

    then using the chain rule and the identity that bnt referred to we get:

    [latex]\frac{dy}{dx}=-sin(x) \frac{1}{\sqrt{1-\cos^2(x)}}[/latex].

    Next, using the trig identity, [latex]1-\cos^2(x)=\sin^2(x)[/latex] we get:

    [latex]\frac{dy}{dx}=-sin(x) \frac{1}{\sqrt{\sin^2(x)}}=-sin(x) \frac{1}{\sin(x)}=-1[/latex].


  • Registered Users Posts: 23 bluesnow


    Thank you very much!


  • Registered Users, Registered Users 2 Posts: 5,629 ✭✭✭TheBody


    Your welcome. Glad I could help!


  • Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    Or you could simply observe that, since sin(y) = cos(x), it follows that y =Pi/2 - x, so dy/dx = -1.

    (Possible additive constant involved, depending on range of x, but it disappears when you differentiate anyway.)


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