Calculus (Derivatives of Inverse Functions)
posted by Mooch .
Suppose f(x) = sin(pi cos(x)). On any interval where the inverse function y = f^1(x) exists, the derivative of f^1(x) with respect to x is:
I've come as far as y = arccos ((arcsin(x))/pi), but I am not certain this is right.

your expression for y is right. so,
Let u = 1/pi arcsinx
(arccos u)' = 1/sqrt(1u^2) u'
= 1/sqrt(1u^2) * 1/pi * 1/sqrt(1x^2)
=  1/sqrt(1 1/pi^2 (arcsinx)^2) * 1/pi * 1/sqrt(1x^2)
=  1/[sqrt(pi^2  (arcsinx)^2)*sqrt(1x^2)]
yummm! gotta love it!
Respond to this Question
Similar Questions

inverse
If f(x)=cosx + 3 how do I find f inverse(1)? 
math
Eliminate the parameter (What does that mean? 
Calculus III
Derivative of Inverse Trigonometric Functions f(x) = sin(arccos(4x)) What is f'(x)? 
Calculus III
Derivative of Inverse Trigonometric Functions f(x) = cos(arcsin(2x)) What is f'(x)? 
Math
Which of the following are inverse functions? 
math (repost)
Which of the following are inverse functions? 
Calculus (Please check my work!)
Suppose that y = f(x) = x^2  4x + 4. Then on any interval where the inverse function y = f1(x) exists, thederivative of y = f1(x) with respect to x is: (Hint: x^2  4x + 4 can be factored and rewritten as "something" squared.) Work … 
Calc AB
Suppose that y = f(x) = x^24x+4 Then on any interval where the inverse function y = f^–1(x) exists, the derivative of y = f^–1(x) with respect to x is: a) 1/(2x4) b) 1/(2y4), where x and y satisfy the equation y=x^24x+4 c)(1/2)x^(1/2) … 
Math
Suppose f(x) = sin(pi*cosx) On any interval where the inverse function y = f –1(x) exists, the derivative of f –1(x) with respect to x is: a)1/(cos(pi*cosx)), where x and y are related by the equation (satisfy the equation) x=sin(pi*cosy) … 
Math
y=arccos(sin(x)), find dy/dx and sketch it's graph(I guess I can do this on wolframalpha after I'm done solving the question). And by arcsin I mean inverse of the expression(written like cos^1(sin(x)), but is not 1/cos(sinx) but the …