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University of North Texas



Problem 52 Medium Difficulty

Find $ y' $ if $ x^y = y^x. $


$y^{\prime}=\frac{\ln y-\frac{y}{x}}{\ln x-\frac{x}{y}}$


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Video Transcript

So if we want to find why, prime, one thing that will help us is if we first take the natural log on each side, just toe put into a little bit easier of a way for us to evaluate the derivative. So first we used to power on each side would be why Natural log of X is equal to X natural law of why. And now if we apply the derivative on each side, we'll need to use product rule in each case, because why is the function we're assuming? That depends on X and natural. Lagerback's is a function that depends on X um, and so first we use product room. So if I take the derivative of why that would be y prime times natural log of X and then plus, so we just have Why times the derivative of natural log of X, which would be one over X and now this is going to be equal to well, if we take the derivative of X, that's just going to be one that we have natural log of why and then plus X times, the derivative of Natural Log, which would be one over wide But then we need to apply chain rule and the derivative of why is just going to be why product. So now what we want to do is get our white crimes on the same side. I'm going to take this term and then just subtract it over. So we have why prime natural log of X eyes equal to or actually minus X y prime over. Why is equal to? And now I'm going to subtract this over. So we first have natural log of y minus and that would be Y Rex. Now we can go ahead and factor out the white prime on the left side. So the UAE prime is even to natural log of X minus X over. Why? And this is eagle toe match log of y, minus y over x. And then we just divide that over a wide prime is in between natural on of why minus y over X all over match log of X minus X over Y um And I guess if you wanted, you could go ahead and clear like the denominators here by multiplying the top and bottom by X on top of bottom. By why? Um but I'm just gonna leave it like this here. So, um, if you get something slightly different, I'm sure if you do a little bit about you can get it into this form here.