Implicit differentiation classic

Question

$ x^t y^m = (x+y)^{m+x} $ prove that $\frac{dy}{dx} = \frac{y}{x} $

I tried and finally got this $\frac{dy}{dx} = \frac{y-1}{1-x} $

Update: I found a solution without using logarithms, $ tx^{t-1}y^m + mx^ty^{m-1} \frac{dy}{dx} = (\frac{(m+t)(x+y)^{m+t}}{x+y}) (1+\frac{dy}{dx})$

$ \frac{dy}{dx} = \frac{tx^{t-1}y^m-\frac{(m+t)(x^ty^m)}{x+y}}{\frac{(m+t)(x^ty^m)}{x+y}-mx^ty^{m-1}}$

$ \frac{dy}{dx} = \frac{tx^ty^m + tx^{t-1}y^{m+1}-mx^ty^m - tx^ty^m}{mx^ty^m+ tx^ty^m - mx^{t+1}y^{m-1}-mx^ty^m}$

$= \frac{x^{t-1}y (ty^m - mxy^{m-1})}{x^ty^{m-1}(ty-mx)}$

$= x^{-1}y = \frac{y}{x}$

Answer 1

We are given $x^ty^m= (x+ y)^{m+ x}$. Since we have the independent variable, x, in the exponent as well as the base, I would start by taking the logarithm of both sides: $t ln(x)+ m ln(y)= (m+ x)ln(x+ y)= mln(x+ y)+ xln(x+ y)$. Now differentiate with respect to x. On the left we get $\frac{t}{x}+ \frac{m}{y}\frac{dy}{dx}$. On the right we get $\frac{n}{x+ y}\left(1+ \frac{dy}{dx}\right)+ ln(x+ y)+ \frac{x}{x+ y}\left(1+ \frac{dy}{dx}\right)$.

So we have $\frac{t}{x}+ \frac{m}{y}\frac{dy}{dx}= \frac{n}{x+ y}\left(1+ \frac{dy}{dx}\right)+ ln(x+ y)+ \frac{x}{x+ y}\left(1+ \frac{dy}{dx}\right)$. Solve that for $\frac{dy}{dx}$.

Answer 2

Your desired result would imply $y=x^p$ for some constant $p$. Substituting that into your main equation would then give $$x^t y^m = x^{t+m p} = (x+y)^{m+x} = (x+x^p)^{m+x}.$$ But the left-hand side is a power law function, and the right-hand side is very much not (for any choice of $m,p$). As such, I see no reason to believe that your main equation in fact implies the desired result.