y = sin(π cosx)
arcsin(y) = π cosx
1/√(1-y^2) = -π sinx dx/dy
but since y=sin(π cosx) this means
1/cos(π cosx) = -π sinx dx/dy
dx/dy = -1/[π sinx cos(π sinx)]
Recall that if y = f(x) and g(x) = f^-1(x) then g'(x) = 1/f'(x)
That is, dx/dy = 1/(dy/dx)
Looks like B to me
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)
1 answer
8 answers