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)
Well, to find the derivative of the inverse function for f(x), we can start by using the formula for the derivative of an inverse function:
(f^−1)'(x) = 1 / f'(f^−1(x))
Now, let's find f'(x):
f(x) = sin(pi*cosx)
Using the chain rule, we have:
f'(x) = (pi*cosx)*(-sin(pi*cosx)) = -pi*cosx*sin(pi*cosx)
Now we can substitute this into the formula for the derivative of the inverse function:
(f^−1)'(x) = 1 / (-pi*cos(f^−1(x))*sin(pi*cos(f^−1(x))))
Since x and y are related by the equation x = sin(pi*cosy), we can replace f^−1(x) with y:
(f^−1)'(x) = 1 / (-pi*cosy*sin(pi*cosy))
And to express it with respect to x, we can replace y with f^−1(x):
(f^−1)'(x) = 1 / (-pi*cos(f^−1(x))*sin(pi*cos(f^−1(x))))
So the answer is e) -1/(siny*cos(pi*cosy)), where x and y are related by the equation x = sin(pi*cosy).
I hope that clears things up! And remember, if math gets too confusing, you can always take a break and juggle some numbers.
To find the derivative of the inverse function y = f^–1(x) with respect to x, we can use the derivative formula for inverse functions.
Let's start by finding the derivative of f(x) = sin(pi*cosx) with respect to x.
f'(x) = d/dx(sin(pi*cosx))
Using the chain rule, we have:
f'(x) = cos(pi*cosx) * (-pi*sinx)
Now, we need to find the inverse function.
Let x = sin(pi*cosy)
To solve for y, we take the arcsine of both sides:
arcsin(x) = arcsin(sin(pi*cosy))
y = cos(pi*arcsin(x))
The inverse function is y = f^–1(x) = cos(pi*arcsin(x)).
To find the derivative of the inverse function, we differentiate both sides with respect to x:
d/dx(y) = d/dx(cos(pi*arcsin(x)))
Using the chain rule, we have:
f^–1'(x) = -pi*sin(pi*arcsin(x)) * (d/dx)(arcsin(x))
The derivative of arcsin(x) is 1/sqrt(1-x^2), so we substitute that in:
f^–1'(x) = -pi*sin(pi*arcsin(x)) * (1/sqrt(1-x^2))
Simplifying, we have:
f^–1'(x) = -pi*sin(pi*arcsin(x)) / sqrt(1-x^2)
Using the trigonometric identity sin(pi*arcsin(x)) = x, we get:
f^–1'(x) = -pi*x / sqrt(1-x^2)
However, the equation we want to satisfy is x = sin(pi*cosy), so we need to express x in terms of y:
sin(pi*cosy) = x
cos(pi*cosy) = sqrt(1-x^2)
cos(pi*cosy) = sqrt(1-sin^2(pi*cosy))
Using the Pythagorean identity sin^2θ + cos^2θ = 1, we have:
cos(pi*cosy) = cos(pi*sin(pi*cosy))
Therefore, we can rewrite the inverse function derivative as:
f^–1'(x) = -pi*x / sqrt(1-sin^2(pi*cosy))
Simplifying further, we have:
f^–1'(x) = -pi*x / cos(pi*cosy)
So the correct answer is option (d) -1/(cos(pi*cosy)).
To find the derivative of the inverse function, we will start by finding the derivative of the original function f(x).
Given: f(x) = sin(pi*cosx)
Let's find the derivative of f(x) using the chain rule:
First, we differentiate the outer function sin(u), where u = pi*cosx, with respect to its inner function u:
d(sin(u))/du = cos(u)
Next, we differentiate the inner function u = pi*cosx with respect to x:
d(u)/dx = d(pi*cosx)/dx = -pi*sinx
Now, we can apply the chain rule:
d(f(x))/dx = d(sin(pi*cosx))/dx = d(sin(u))/du * du/dx = cos(u) * (-pi*sinx) = -pi*sinx*cos(pi*cosx)
Therefore, the derivative of f(x) is -pi*sinx*cos(pi*cosx).
To find the derivative of the inverse function, we can use the formula:
dy/dx = 1 / (dx/dy)
where dy/dx is the derivative of the inverse function and dx/dy is the derivative of the original function f(x).
In this case, we want to find dy/dx, where x and y are related by the equation x = sin(pi*cosy).
Rearranging the equation, we have y = cos^(-1)((arcsin(x))/pi).
Taking the derivative of both sides with respect to x, we get:
dy/dx = d(cos^(-1)((arcsin(x))/pi))/dx
Now, using the chain rule, we differentiate the outer function cos^(-1)(u) with respect to its inner function u:
d(cos^(-1)(u))/du = -1 / sqrt(1 - u^2)
Next, we differentiate the inner function u = (arcsin(x))/pi with respect to x:
d(u)/dx = d((arcsin(x))/pi)/dx = (1 / pi) * d(arcsin(x))/dx
To find d(arcsin(x))/dx, we can use the fact that d(arcsin(u))/du = 1 / sqrt(1 - u^2). In this case, u = x.
Therefore, d(arcsin(x))/dx = 1 / sqrt(1 - x^2).
Substituting these values back into the equation, we have:
dy/dx = -1 / sqrt(1 - ((arcsin(x))/pi)^2) * (1 / pi) * (1 / sqrt(1 - x^2))
Simplifying this expression, we get:
dy/dx = -1 / (pi * sqrt(1 - ((arcsin(x))/pi)^2) * sqrt(1 - x^2))
Therefore, the correct answer is (b) -1/(pi*sinx*(cos(pi*cosx))), where x and y are related by the equation x = sin(pi*cosy).
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