How would you go about differentiating this question, say with respect to x? Find y'. This is definitely giving me some issues. Please Register or Log in to view the hidden image!

Damn... tough one x^y=y^x y=x^(y/x) by analysis this is true if y/x=1 (although I'm not sure how to prove it besides plotting a contour of x^y-y^x=z) so this would simplify to x=y x'=1

But lethe, I did that, but you still end up with a y in the other side. Is that acceptable? (Pardon me, I'm second year and I'm not fond of Calculus.)

x^y = y^x is a transendental equation, so i would expect there to be a y on both sides of the equation. if you can t solve the original equation for y, i don t see why you should be able to solve it in the derivative equation.

Thank you. No it isn't a homework problem, just something one of my friends threw at me to make me think.

For what it's worth, here's what Maple thinks about it: Please Register or Log in to view the hidden image!

To answer the question that was asked earlier, yes, it quite acceptable to write dy/dx in terms of both x and y, in fact it might be NECESSARY. Of course, theoretically, one should be able to solve for y as a function of x and replace it but the whole point of "implicit differentiation" is that it is applicable when you CAN'T solve for x. Implicit differentiation is applicable even when y is NOT a function of x. A simple example is : x<sup>2</sup>+ y<sup>2</sup>= r<sup>2</sup>. While it is easy to solve for y, it's double valued: we have to choose either + or -. On the other hand, differentiating the equation gives 2x+ 2yy'= 0 so y'= -x/y. Here it wouldn't make sense to ask for the derivative at a specific value of x: there may be two different values of y for a single value of x- you have to give both x and y.