Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $y'$.

share|cite|improve this answer

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.

share|cite|improve this answer
+1 for 'it's easier to derive it than to remember it'. – Lord_Farin Apr 23 '13 at 10:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.