Question

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)

b)-1/(pi*sinx*(cos(pi*cosx))), where x and y are related by the equation x=sin(pi*cosy)

c)-1/(pi*siny*(cos(pi*cosy))), where x and y are related by the equation x=sin(pi*cosy)

d)-1/(cos(pi*cosy)), where x and y are related by the equation x=sin(pi*cosy)

e)-1/(siny*cos(pi*cosy)), where x and y are related by the equation x=sin(pi*cosy)

Answer 1

Inverse of sin(pi*cosx) = cos^(-1)((sin^(-1)(x))/pi)

Derivative of cos^(-1)((sin^(-1)(x))/pi) is

-1/(sqrt(1-x^2) sqrt(pi^2-sin^(-1)(x)^2))

Answer 2

derivative: sin(pi*cosx) = -pi sin(x) cos(pi cos(x))

Answer 3

It was B

Answer 4

If anything it wasn't B because I have put in the answer for B and it didn't work. It must be A, C, D, or E instead.

Answer 5

for others who need help the answer is -1/(pi(siny(cosy) the with respect to x messed me up.

Answer 6

Which one is that? I don't think that's an option

Answer 7

Its c

Answer 8

Its -1/pi(sinycos(picosy) where x and y are related by the equation x=sin(picosy)