Question

If a function f(x) has a slope m, then its inverse function f^-1(x) has a slope 1/m. So how come the derivative of ln(x) is not e^-x ?

Answer 1

I'm pretty sure the slope of the inverse function \(f^{-1}(x)\) is \(-1/m\) and not \(1/m\)

Answer 2

Well if you have y=2x the inverse function is y=x/2 isn't it?

Answer 3

Gimme some time .. I am not really sure -.-

Basically the way to find out is to note the fact that \(f(x)\) and \(f^{-1}(x)\) are symmetric about the line y = x.

Answer 4

Sure, that's why I'm saying this haha.

Answer 5

So like in the simple case we have: \[\LARGE f = 2x+3 \ \ \ \ f^{-1}=\frac{1}{2}x+\frac{3}{2} \\ \LARGE f' = 2 \ \ \ \ (f^{-1})' =\frac{1}{2}\] So obviously for lines \[\LARGE f' = \frac{1}{(f^{-1})'}\] that will hold we could just run the whole thing through with m and b where 2 and 3 were.

But it doesn't work always it seems. Obviously the derivative of e^x is not the reciprocal of the derivative of lnx.

Answer 6

Also, this is just a hunch, but I think the relation regarding the slopes is valid only at the point at which the function and it's inverse intersect. (or if it's linear).

For non-linear functions, it's only valid at the point of intersection.

If they don't intersect, then it's never valid (atleast for the same x).

Answer 7

*Never* in the last line is a pretty strong word ... Don't take it literally :P

Answer 8

No you're wrong.

Answer 9

The inverse function is a reflection across the line y=x. They have inverse slopes of each other at every point.

Answer 10

correct me if i'm wrong but:
"The inverse function is a reflection across the line y=x. They have inverse slopes of each other at every point \(\text{in a rotated axis aligned with the line y-x}\)"

Answer 11

I'll be back in an hour or two .. Hopefully I'll think up of something concrete till then.

ttyl

Answer 12

Uhh, this is embarassing .. I would suggest anyone looking at my posts in this thread to IGNORE EVERYTHING.....

(I have a strong inclination to delete all these pots, but i'll keep em for the sake of @Kainui so that it doesn't seem that he's been talking to himself) ^.^

Answer 13

http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation

Nah I know the answer all along I was just playing. This should make it clear. ;P

Answer 14

I was going to say that the relation between the inverse functions' derivatives has something to do with them being onto.

Clearly, \(f,g:\mathbb{R}\to\mathbb{R}\), \(f(x)=\ln x\) and \(g(x)=e^x\) are not onto.

Answer 15

(¬д¬。)

Answer 16

@Kainui Those are two somewhat unrelated statements.
Here's how I see your question: if a bird has wings and can fly, who do dolphins swim?

Answer 17

Things don't map so easily.

Answer 18

\[\LARGE y(x)=e^x \ \ \ \ x(y)=lny \\ \LARGE y'(x) = e^x \ \ \ \ x'(y)=\frac{1}{y}\]

so we know \[\LARGE y'(x)=\frac{1}{x'(y)}\] is clearly true since \[\LARGE e^x=\frac{1}{1/y}\]

It's just a trick question.

Answer 19

What are you talking about people? @primeralph @SithsAndGiggles

Answer 20

In my defense, it was about 3 am when I posted that...

Answer 21

=P