Posted by **Anonymous** on Thursday, March 15, 2012 at 2:10pm.

Suppose that y = f(x) = x^2 - 4x + 4.

Then on any interval where the inverse function y = f-1(x) exists, thederivative of y = f-1(x) with respect to x is:

(Hint: x^2 - 4x + 4 can be factored and rewritten as "something" squared.)

Work thus far:

Okay, so I know that I can factor y to equal y = (x - 2)^2

Next, I can solve for the inverse function:

x = (y - 2)^2

sqrt(x) = y - 2

y = sqrt(x) + 2

Then, I can find the derivative:

y' = (1/2)x^(-1/2)

Now, my concern is that the question also hints to an implicit solution to this problem. Working out the problem implicitly, I get:

x = y^2 - 4x + 4

1 = 2y(dy/dx) - 4(dy/dx)

dy/dx = 1 / (2y - 4) where x and y satisfy the equation y = x^2 - 4x + 4.

Would this answer be correct, as well? Or should it be "where x and y satisfy the equation x = y^2 - 4y +4"?

- Calculus (Please check my work!) -
**Reiny**, Thursday, March 15, 2012 at 3:39pm
from y = f(x) = x^2 - 4x + 4

forming the inverse consists of interchanging the x and y variables, so the inverse is

x = y^2 - 4y + 4

(you had -4x instead of -4y, but apparently worked it as -4y )

so 1 = 2y dy/dx - 4dy/dx

dy/dx(2y-4) = 1

dy/dx of the inverse is 1/(2y-4)

but your restriction would be y ≠ 2

## Answer this Question

## Related Questions

- Calculus (Derivatives of Inverse Functions) - Suppose f(x) = sin(pi cos(x)). On ...
- Calc AB - Suppose that y = f(x) = x^2-4x+4 Then on any interval where the ...
- Calculus, check my answers, please! 3 - Did I get these practice questions right...
- pre calculus - Find the inverse of the function below. Graph the function below ...
- calculus - Consider the function f(x)=3x^2 – 5x on the interval [-4,4]. Find the...
- CALCULUS! - Consider the function f(x)=4x^3–2x on the interval [–2,2]. Find the...
- Calculus - The function f(x)=(x^4)-(10x^3)+(18x^2)-8 is continuous on the closed...
- Calculus Please help! - Consider the function f(x)=4sqrt(x)+4 on the interval [2...
- College Algebra - Find the inverse of the function below. Graph the function ...
- Calculus - Solve the following initial value problem and determine the interval ...