Discussion between Jens Renders and Vivaan Daga

Feeds

11:14

A: Why is this theorem about derivatives true? $\frac{dy}{dx}= \frac{1}{dx/dy}$

This question has some good answers already, but I want to point out that the intuition from abusing the notation can lead to a proof directly. Just using the limits behind the derivative notation works: $$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta y \to 0} \fr...

Vivaan Daga

If f inverse is not differentiable then this fails is there an example of such

Please answer my edit

Jens Renders

@VivaanDaga What do you mean an example? If $f^{-1}$ is not differentiable then the theorem statement does not make sense. It contains the derivative of $f^{-1}$: $\frac{1}{dx/dy} = 1/(f^{-1})'(y)$ by definition.

Vivaan Daga

the theorem will fail but can the derivative still exist? i would not think so because you need to express x as a function of x in dx/df ? am i correct?

Jens Renders

@VivaanDaga Can what derivative still exist?

Vivaan Daga

dx/df if f inverse does not exist/differentiable

Jens Renders

11:14

@VivaanDaga but $\frac{dx}{dy}$ is by definition the derivative of $x = f^{-1}(y)$ to $y$.

Vivaan Daga

Yes so if there is no inverse/inverse is not differenetiable then it wont exist am i correct?

so derivatives like dx/dc where c is a constant function or dx/d|x| have no meaning is that correct?

Jens Renders

You are asking, if f has no inverse, can its inverse have a derivative?

that is a nonsensical question

you can not take a derivative of a function that does not exist

in other notation you are asking

if f is a function such that there is no inverse f^-1

does (f^-1)' exist?

Vivaan Daga

But how is dx/df same as taking derivative of f inverse exactly

Jens Renders

dx/df is wrong notation

if f is a function of x

for a function f(x), its derivative to x is denoted df(x)/dx

or df/dx

and if y = f(x)

then we can write dy/dx

so we can only write dx/dy if x is a function of y

Vivaan Daga

But if g(f(x))=x then f is function of x and then g will have to be f inverse so if f inverse does not exists /differentiable then dx/df has no meaning is that correct

Jens Renders

11:23

indeed

and dx/dy has also no meaning then

the notation dx/dy means, write x as a function of y and then take derivative to y

so if we start from y = f(x)

to have x as a function of why we need f^-1 to solve for x

x = f^-1(y)

and then we take derivativ to y

so f^-1(y) has to be differentiable

otherwise dx/dy has no meaning for a function y = f(x)

Vivaan Daga