Find (f −1)'(a). f(x) = 3x^3 + 2 sin x + 3 cos x, a = 3
I don't know how to solve this question
f'(x) = 9x^2 + 2cosx
f(0) = 3
so (f-1)'(3) = 1/f'(0) = 1/2
ah ty!
To find the derivative of the inverse of a function at a specific value, you can follow these steps:
Step 1: Find the derivative of the original function, f(x).
Step 2: Swap x and y in the equation f(x) = a and solve for y to get the inverse function, f^(-1)(y).
Step 3: Differentiate f^(-1)(y) implicitly with respect to y.
Step 4: Substitute the value of the function output a into the derivative you obtained in step 3.
Now, let's go through each step to find (f^(-1))'(a) for the given function f(x) = 3x^3 + 2 sin(x) + 3 cos(x) at a = 3.
Step 1: Find the derivative of f(x):
f'(x) = 9x^2 + 2 cos(x) - 3 sin(x).
Step 2: Swap x and y in the equation f(x) = a:
x = 3y^3 + 2 sin(y) + 3 cos(y).
Step 3: Differentiate implicitly with respect to y:
1 = 9y^2 + 2 cos(y) - 3 sin(y).
Step 4: Substitute a = 3 into the derivative from step 3:
1 = 9(3)^2 + 2 cos(y) - 3 sin(y).
Simplifying this equation will give you the value of y at which the derivative is equal to 1.