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March 29, 2015

March 29, 2015

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

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 3:39pmfrom 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

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