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Use implicit differentiation to find $\frac{\mbox{dy}}{\mbox{dx}}$ if $$x^my^n = (x+y)^{m+n}$$

Differentiation both sides with respect to $x$:

$$mx^{m-1}y^n + x^mny^{n-1}y' =(m+n)(x+y)^{m+n-1}(1 + y')$$ $$y' = \frac{(m+n)(x+y)^{m+n-1} - mx^{m-1}y^n}{x^mny^{n-1}-(m+n)(x+y)^{m+n-1}}$$ $$y'= \frac{nxy-my^2}{nx^2-mxy}$$ after using given $x^my^n = (x+y)^{m+n}$. The answer given to me is $\frac{y}{x}$. So, it seems $y'$ can be simplified even more.

How do we do this?

Thanks

MangoPizza
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3 Answers3

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You just didn't finish.

$\begin{array}\\ y' &= \frac{nxy-my^2}{nx^2-mxy}\\ &= \frac{y(nx-my)}{x(nx-my)}\\ &= \frac{y}{x}\\ \end{array} $

Note that if $nx=my$ you can not take this final step.

marty cohen
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Simpler way: Take logarithms (base e unless specified). $$m\log x+n\log y=(m+n)\log (x+y)$$ Now differentiate: $$\frac mx +\frac ny\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)$$ Therefore $$\frac mx-\frac{m+n}{x+y}=\left(\frac{m+n}{x+y}-\frac ny\right)\frac{dy}{dx}$$$$\implies \frac{my-nx}{x(x+y)}= \frac{my-nx}{y(x+y)} \cdot \frac{dy}{dx}$$ so $$\frac{dy}{dx}=\frac yx.$$

insipidintegrator
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  • How does this work? What if $x$ or $y$ is negative, then we can't take logarithm... – MangoPizza Aug 14 '22 at 05:23
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    Hi, I deleted my earlier comment because I thought these posts could clarify better: https://math.stackexchange.com/questions/415141/is-there-a-problem-when-defining-exponential-with-negative-base and https://math.stackexchange.com/questions/3121464/y-kx-when-k-is-negative. The problem is when $m,n \notin \mathbb Z$, not when $m,n<0$. – insipidintegrator Aug 14 '22 at 05:30
  • Thank you, the problem is when $x, y$ is negative and $m, n$ are some non-integers. But what if I modify the question such that $m, n$ are guaranteed to be integers? Then $x, y$ can take negative values, then log can't be taken. – MangoPizza Aug 14 '22 at 06:19
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    Yes, then it will be better to use the other methods. – insipidintegrator Aug 14 '22 at 06:20
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For any homogeneous relation of $x$ and $y$, we always have $$\frac{dy}{dx}=\frac{y}{x}.$$ Because the relationship is always satisfied by $y=vx$ meaning $$\frac{dy}{dx}=v=\frac{y}{x}.$$ The relation $$x^my^n=(x+y)^{m+n}$$ is a homogeneous function of degree $m+n$.

For $x^2+xy+y^2=0$ and $4x^2y+5y^3+6x^3-7xy^2=0.$ After finding you have to use back the relationship to get $\frac{dy}{dx}=\frac{y}{x}$

Z Ahmed
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