How to find the derivative dy/dx for y^x=x^y (calculus 1 tutorial logarithmic differentiation)

We will find the derivative of y^x=x^y, namely, we will find dy/dx by using implicit differentiation and logarithmic differentiation for the equation y^x=x^y. This is a very classic calculus 1 implicit differentiation example!

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[leickrobinson]

It seems like you can simplify it a little bit further through substitution using the ln of the original equation so that you get a numerator that is only in terms of y and the denominator that is only in terms of x. Plus, you can then factor out Y squared and X squared in the numerator and denominator, respectively, and separate them from the ln terms.

[tigerinthejungle_]

We can use this also:-
dy/dx = partial derivative of f w.r.t x / partial derivative of f w.r.t y, here f = xlny - ylnx

[Marcus-ym]

I was looking for another way to solve it, thank you

[alfiekern]

Never imagine a guy on YouTube helps me doing my homework while holding a pokemon ball

[FrogworfKnight]

I did thing in a way that used the quotient rule instead, so the answer looks different.
y^x=x^y
ln(y^x)=ln(x^y)
xln(y)=yln(x)
ln(y)/y=ln(x)/x
(d/dx)(ln(y)/y)=ln(x)/x)

[stephenbeck]

Put a dy/dx at the very beginning as a coefficient on the left side. Can you separate variables and integrate?

[sepdronseptadron]

is it possible to find the derivative in terms of only x?

[dfsgjlgsdklgjnmsidrg]

x*ln(y) = y*ln(x) judt divide both divide both sides by x*y and get ln(y)/y = ln(x)/x therefore x=y and therefore easy peasy

[vishwas]

Hi !
Can you please make a calculus course from beginning.

[amilanissanka]

Why do we have to take the ln of both sides before differenciating? Doesn't that change the answer of the original x^y = y^x?

[RudeBuddha]

Please compare the rates of both functions

[PLAYWORD]

Hey i did someone similar to that.
I was in my boring room and i thought hmm X^y=X/2

If x is any real value how to find and y that if we elevate X to the power y we get X/2?

[ProfessionalRacist]

Now, do this while taking all restrictions for x and y.

[DeJay]

Just a reminder that x^y=y^x is NOT a function