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

Рет қаралды 11,279

Күн бұрын

Пікірлер: 39

@NoisyKoel 2 ай бұрын

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)

@bprpcalculusbasics 2 ай бұрын

Yes you are right. Thanks for pointing that out!

@NoisyKoel 2 ай бұрын

@bprpcalculusbasics my pleasure!! 😁 Thank you for your thorough lessons!! 😃

@markcbaker 2 ай бұрын

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

@stupidteous 2 ай бұрын

he did simplify in the last video. he was just showing how it could be done the other way to clear up the viewer's question at the beginning. it should be simplified though.

@mtaur4113 2 ай бұрын

I am also wondering if lny/y = lnx/x has nontrivial solution. y=x is a solution curve. But there is another one, G(t)=ln(t)/t has a maximum at t=e and has pairs of positive values for t_1>e and 1

@giovannimariatanda9251 2 ай бұрын

Bravo @markcbaker. Pure io penso sia più semplice e forse più elegante fare come dici. Così ho infatti proceduto in prima istanza. Ad ogni modo onore anche a BRP per il suo prezioso contributo. Virtute duce, comite fortuna hostes vicisti. (Cicero)

@stupidteous 2 ай бұрын

@@mtaur4113 i think he did make a video on it, x^y = y^x

@intothedarkness6242 2 ай бұрын

same,i ended up with lny - yx^-1 / lnx - xy^-1

@chakwowu6825 2 ай бұрын

I like the X'mas tree behind you.

@profjaildervicente 2 ай бұрын

Parabéns, muito bom o seu conteúdo...

@issssse 2 ай бұрын

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 ❤

@henkhu100 2 ай бұрын

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

@chaosredefined3834 2 ай бұрын

That's not a problem. We can also use this approach to find the derivative of a circle (x^2 + y^2 = r^2), despite the fact that, for any x such that -r < x < r, there will be two values of y that work. Things get a bit loose when working with derivates.

@TheEternalVortex42 2 ай бұрын

Look up the implicit function theorem. It guarantees we can write this function as differentiable y = f(x) or x = g(y) everywhere.

@henkhu100 2 ай бұрын

@@TheEternalVortex42 The implicit function theory is not always applicable. It is valid under certain conditions. suppose we have the following relation in the real number system: (x+lny)sqrt(-(x-3)^2)=x^2-y You can try to find dy/dx but does it have any meaning? The only x value in the domain is x=3 (with y=9). So the graph is a single point. has dy/dx any meaning despite the fact that you can find an expression for it? You can even define a relation where there is no graph at all: (x+lny)sqrt(-(x-3)^2)=x^2-y + sqrt(-(y-3)^2) but you can apply implicit differentiation giving a meaning less result. So it is always important to know that a given relation between x and y results in a differentiable function. You write: "It guarantees we can write this function as differentiable y = f(x) or x = g(y) everywhere." But are we sure that the given relation in the video complies with the conditions to use the principle of implicit differentiation?

@hydra-f9h 2 ай бұрын

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

@kevinpior5265 2 ай бұрын

Because y is a function of x here

@ve4rexe 2 ай бұрын

because y is a function of x, in x

@kb27787 2 ай бұрын

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?

@chaosredefined3834 2 ай бұрын

@TheEternalVortex42 2 ай бұрын

This is called an implicit equation. By the implicit function theorem we have some conditions under which an implicit equation gives you one or more (differentiable) functions. In that case the approach of implicit differentiation is correct for finding the derivatives of those functions. For example the simpler implicit equation x^2 + y^2 = 1 gives four functions y = sqrt(1-x^2) and x = sqrt(1-y^2) and the flipped versions. But in post cases you cannot write them down explicitly.

@priscilas24 2 ай бұрын

Now I can understand lot better 😊

@miguelaphan58 2 ай бұрын

. a most usefull lesson !!

@arcangyal2269 2 ай бұрын

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

@TheBlueboyRuhan 2 ай бұрын

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

@andirijal9033 2 ай бұрын

Explisit function ?

@Mediterranean81 2 ай бұрын

I did it using implicit differentiation

@Mediterranean81 2 ай бұрын

y*ln x = x ln y y/ln y = x/ln x ln y / y = ln x /x y’ (1-ln y) /y^2 = (1-ln x)/x^2 y’ = (1-ln x)y^2 /(1-ln y)x^2

@jasonryan2545 2 ай бұрын

@@Mediterranean81Is there any reason we can manipulate the fraction even more by increasing their power to the negative one? It seems a bit too much like steamrolling through....

@perost1227 2 ай бұрын

nice

@chrisglosser7318 2 ай бұрын

I wrote everything in terms of logs

@KeaneMbae 2 ай бұрын

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