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Determine the derivative.$$e^{x^{2}}-e^{y^{2}}=y^{2}-x^{2}$$

$\frac{x\left(1+e^{x^{2}}\right)}{y\left(1+e^{y^{2}}\right)}$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 4

The Derivative of the Exponential Function

Missouri State University

Oregon State University

Harvey Mudd College

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eso This problem requires implicit differentiation because as you look at it, we have e to the X squared minus e to the y squared. Definitely not a function, uh, equals y squared minus x squared Eso What do I mean by implicit is you take the derivative of each side. So first of all, the derivative of E to the X squared is you leave e to the X squared alone, but then times the derivative of export, which is two X and now the next one is actually the same thing. Except it's changeable inside of change rule, which means that you have toe also say implicitly that you have to do D y d s eso after you take the derivative of X, you know, way have to do that for the derivative of y the Y dx. Same thing over here we have two y times d y d x on then the directive of export is just two X you don't have to worry about. I guess you could if you want to write dx DX because that's your independent variable. But anything over itself is just one. So it's zero waste of time. Um Now, what I would do next is get my dear, why the X is on the same side. And so what I would do is leave this over here because I'm thinking, Well, this is a negative term, so I can add, um and multiplication is communicative. So I'm just moving this piece in front to I d y dx, um, e to the y squared power. And you know this color code. So again, multiplication is communicative, so I could move to X in front of the E to the X squared. I'm gonna add two X to the left side. Eso Everything's positive now, which is kind of nice. But anyway, their next step is to find the greatest common factor, which is the greatest common factor here on you don't even have to do the greatest contact. We could just leave everything else. I'm sure the answer key probably get the answer. He definitely divides everything by two because it's a common factor on everything here. Yeah, I know everything by two Onda. While I'm looking at it looks like they also factor out of why on this side on your left, with what one plus e to the y squared since I did it there. Amounts would do that on this side that you could factor out in X E to the X squared plus one. And now just think of this is a giant multiplication problem. So how do we solve? Multiplication is through division. So I'm not messing with that left side at all. I just have to divide why? And I don't know why I'm rewriting it. Just I guess I have a habit. And this would be your answer in additions community so you can switch it around.

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