Maths Differentiation Problem

bluesnow

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.

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.

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].

bluesnow

Thank you very much!

TheBody

Your welcome. Glad I could help!

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.)