Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi all! We have been experiencing an issue on site where threads have been missing the latest postings. The platform host Vanilla are working on this issue. A workaround that has been used by some is to navigate back from 1 to 10+ pages to re-sync the thread and this will then show the latest posts. Thanks, Mike.
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Maths Differentiation Problem

  • 28-09-2014 5:49pm
    #1
    bluesnow
    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.
    Tagged:


Comments

  • #2
    bnt
    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



  • #3
    TheBody
    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].


  • #4
    bluesnow
    Registered Users Posts: 23 ✭bluesnow


    Thank you very much!


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


    Your welcome. Glad I could help!


  • #6
    MathsManiac
    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.)


  • Advertisement
Advertisement