William Boggs

2022-01-22

Implicit differentiation question
Given: $\frac{y}{x-y}={x}^{2}+1$

Pansdorfp6

Expert

$\frac{y}{x-y}={x}^{2}+1$
You claim that ${y}^{\prime }=2x$
so that $y={x}^{2}+C$
This means $\frac{{x}^{2}+C}{x-{x}^{2}-C}={x}^{2}+1$
This is absqrt, since the quotient of two second degree polynomials can't be a second degree polynomial. In fact you get two non vanishing terms which are off.
I don't understand what your procedure is, also. I would proceed as follows:
$\frac{y}{x-y}={x}^{2}+1$
$\frac{d}{dx}\left(\frac{y}{x-y}\right)=\frac{d}{dx}\left({x}^{2}+1\right)$
$\frac{{y}^{\prime }\left(x-y\right)-\left(1-{y}^{\prime }\right)y}{{\left(x-y\right)}^{2}}=2x$
$\frac{{y}^{\prime }x-y}{{\left(x-y\right)}^{2}}=2x$
${y}^{\prime }x=2x{\left(x-y\right)}^{2}+y$
${y}^{\prime }=2{\left(x-y\right)}^{2}+\frac{y}{x}$

Barbara Meeker

Expert

An explicit approach:
Rewrite as $y=\left(x-y\right)\left({x}^{2}+1\right)$, and factor out y to get $y=\frac{{x}^{3}+x}{{x}^{2}+2}$, This is straightforward to differentiate, yielding $\frac{dy}{dx}=\frac{{x}^{4}+5{x}^{2}+2}{{\left({x}^{2}+2\right)}^{2}}$.

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