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)
Its c
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.
for others who need help the answer is -1/(pi(siny(cosy) the with respect to x messed me up.
Its -1/pi(sinycos(picosy) where x and y are related by the equation x=sin(picosy)
Which one is that? I don't think that's an option
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))
derivative: sin(pi*cosx) = -pi sin(x) cos(pi cos(x))
To find the derivative of the inverse function, we can use the formula:
dy/dx = 1 / (df/dy)
First, let's find df/dy:
f(x) = sin(pi*cosx)
To find df/dy, we need to express f(x) in terms of y. We can do this by replacing x with y in the equation x = sin(pi*cosy):
x = sin(pi*cosy)
Now, let's solve for sin(pi*cosy):
sin(pi*cosy) = x
Next, let's solve for cosy:
cosy = arcsin(x) / pi
Now, substitute this expression for cosy back into f(x):
f(x) = sin(pi*cos(arcsin(x)/pi))
Simplifying further:
f(x) = sin(arcsin(x))
Since arcsin(x) gives us an angle, and sin(arcsin(x)) gives us the value of x, the expression simplifies to:
f(x) = x
So, df/dy = 1.
Now, we can find the derivative of the inverse function dy/dx:
dy/dx = 1 / (df/dy) = 1.
Therefore, the correct answer is (d)-1/(cos(pi*cosy)), where x and y are related by the equation x = sin(pi*cosy).