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Find $y'(x)$ for $y=y(x)$ if:

a) $\sin(xy) -e^{xy}-x^2y=0$

b) $x^y+y^x=0$

So the formula for these types of functions is $dx/dy = -F_x(x,y)/F_y(x,y)$. How to apply this?

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The equation $x^y+y^x=0$ has no solutions in real variables $x$, $y$, let alone a solution of the form $x\mapsto y(x)$. – Christian Blatter Dec 22 '13 at 15:10

Lets do what @Lord_Farin pointed above.

a) If $\sin(xy) -e^{xy}-x^2y=0$ while $y=y(x)$, then we are looking for $y'$ and so we should do:$$\left(\sin(xy)\right)_x -\left(e^{xy}\right)_x-\left(x^2y\right)_x=0$$ Or $$\left(y\cos(xy)+xy'\cos(yx)\right)-(ye^{xy}+xy'e^{xy})-(2xy+x^2y')=0$$ And so: $$y'=\frac{-y\cos(xy)+ye^{xy}+2xy}{x^2-xe^{xy}+x\cos(xy)}$$

b) $x^y-y^x=0$. Under some conditions for $x$ and $y$ for example $x>0,y<0$; then $$x^y=-y^x\to y\ln x=x\ln(-y)\to y\ln(x)-x\ln(-y)=0$$ and as we did for case a above, we have: $$\frac{y}{x}+y'\ln(x)-\ln(-y)-\frac{xy'}{y}=0$$ Now try to factor the latter statement for finding $y'$.

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I would use implicit derivation. If you have a function $f(x,y) = 0$ and $ y = y(x)$ then $ \frac{ \partial f(x,y)}{ \partial x} = \frac{ \partial f(x,y)}{ \partial x} + \frac{ \partial f(x,y)}{ \partial y}\frac{ \partial y}{ \partial x} = 0 $, i.e $$ \frac{ \partial y}{ \partial x} = - \frac{ \frac{ \partial f(x,y)}{ \partial x}}{\frac{ \partial f(x,y)}{ \partial y}} $$ and as you correctly pointed out this gives the formula you posted. It much easier to derive the formula, than to remember it. So for the first one you have $ f(x,y) = \sin(xy) - e^{xy} -x^2y$, with $\frac{ \partial f(x,y)}{ \partial x} = y \cos(xy) - y e^{xy} - 2xy$ and $\frac{ \partial f(x,y)}{ \partial y} = x \cos(xy) - x e^{xy} - x^2$. Now I think you can complete the first one by just inserting this in the formula.

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+1 for 'it's easier to derive it than to remember it'. – Lord_Farin Apr 23 '13 at 10:20

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