If x^y=y^x, then dy/dx=?

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I love your videos!! I am picking up Calculus to do my Masters soon. By the way, at around 7:32, shouldn't the whole thing be multiplied by xy / x^y in the numerator AND the denominator?

Because at 7:32, I think you accidentally wrote the multiplying denominator as y^y instead of x^y. (I.e. xy / y^y when it should be xy / x^y)

NoisyKoel

You first have to find out that y is a differentiable function of x if x^y=y^x before you can use the expression dy/dx
See also an earlier reaction that there are two values for y if x=2

henkhu

Parabéns, muito bom o seu conteúdo...

profjaildervicente

1:57 2:50 Why is y Not a constant, our only variable to respect when differenciating is x

hydra-fh

Does this exist? Let's say x = 2 there will be 2 Y values (2 and 4) that fulfill this condition. So this is not a function in the strictest sense as one input can give multiple outputs. I imagine the graph would look like the straight line with slope of 1 (x=y) and then something else. Can we take the derivative of something that is not a function?

kb

dy/dx = -Fx / Fy makes it so much faster and easier to do

TheBlueboyRuhan

Some series approach infinity pretty quickly, and some very slowly 🐌 (just watched your harmonic series video). But which series approach infinity the slowest?

Love the videos ❤

issssse

Why not simplify to y*ln x = x*ln y before implicit differentiation

markcbaker

Can we use dy/dx = - partial x / partial y?

arcangyal

I did it using implicit differentiation

Mediterranean

I like the X'mas tree behind you.

chakwowu

I always go to the most easy one, the number 1 😂😂😂😂 the other ones are for crazy people❤😅

cesarluis

Your really smart and great at calculus why don't you try using all this math skills in writing a good program with a functional programming language e.g. Haskell/APL maybe you are also good at that

KeaneMbae

The original equation implies that
Y equals X.
... Is it not so?
If it is so, then dy/dx is 1.
Simplex sigillum veri.

notsoancientpelican

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