Derivative of the inverse of a function

[yurkler]

Homework Statement

y=x+(1/x) at y=17/4

Homework Equations

The Attempt at a Solution

y^-1: x=y+(1/y)

differentiate: 1=y'+ln(y)y'
1=y'(1+ln(y))
y'=1/(1+ln(y))

put that over 1: 1+ln(y)

plug in y: 1+ln(17/4)
=approximately 2.447

 

[Mark44]

the function is y=x+(1/x) at y=17/4

attempt:
y^-1: x=y+(1/y)

Here, by switching letters, you have x = f(y). f is the same function as above, except that its argument is now called y.

differentiate: 1=y'+ln(y)y'

You differentiated with respect to x, so what you have above is 1 = f'(y)*dy/dx

1=y'(1+ln(y))
y'=1/(1+ln(y))

Now you have dy/dx = 1/f'(y)

put that over 1: 1+ln(y)

What you're saying and what you're doing are two different things. If you put anything over 1, you get exactly the same thing.

What you did is take the reciprocal. Why?

plug in y: 1+ln(17/4)
=approximately 2.447

You switched letters, so what was the old y value (17/4) is now an x value. The question is, what is the new y value?

The answer I get is ~.42

 

[cepheid]

The inverse function is NOT given by x = y + 1/y

To see this, just use a test value. e.g., if x = 2, then y(2) will give you some value. If you then plug this value into your "inverse" function (i.e. x(y(2)), you should get back 2. But you don't, because:

y(2) = 2 + 1/2 = 5/2

x(5/2) = 5/2 + 2/5 = 25/10 + 4/10 = 29/10

29/10 is not equal to 2. Since you didn't get back what you started with, your "inverse" function x(y) must be wrong.

To get the actual inverse function, solve the equation to find x in terms of y. In other words, solve the equation for x. Hint: you'll find that the function does not have a unique inverse.

 

[cng99]

The answer I get is ~.42

42!