Calculus (Please check my work!)
posted by Anonymous .
Suppose that y = f(x) = x^2  4x + 4.
Then on any interval where the inverse function y = f1(x) exists, thederivative of y = f1(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
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(2y4) = 1
dy/dx of the inverse is 1/(2y4)
but your restriction would be y ≠ 2 
Calculus (Please check my work!) 
Ashley
^thats wrong. I got it wrong. sorry.
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