# Finding derivatives using implicit differentiation

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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Hello, welcome to Math.SE. In general, it will help if you add some of your thoughts on the problem, so that we don't write things you already know or fly over your head. If this is homework, please add the homework tag (don't worry, we'll still help you). Finally, for some basic information about writing math at this site see e.g. here, here, here and here. – Lord_Farin Apr 23 '13 at 9:57
If you are a beginner you can try codecogs.com/latex/eqneditor.php <-this sites which will automatically generate the code using a graphical interface p – fermesomme Apr 23 '13 at 10:03
It should be $dy/dx = -F_x(x,y)/F_y(x,y)$ – Mhenni Benghorbal Apr 23 '13 at 10:18
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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@B.S. needs another UV +1 – Amzoti Dec 23 '13 at 14:27

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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Nice, I removed the "almost" ;) – Henrik Finsberg Apr 23 '13 at 10:19
+1 for 'it's easier to derive it than to remember it'. – Lord_Farin Apr 23 '13 at 10:20