Question

Given the equation y ln (x^2 + y^2 + 8) = 3​, evaluate dy/dx. Assume that the equation implicitly defines y as a differentiable function of x.

Answer 1

$y\ln(x^2+y^2+8)=3$

Differentiate both sides with respect to $x$:

$\dfrac{\mathrm dy}{\mathrm dx}\ln(x^2+y^2+8)+y\dfrac{\frac{\mathrm d}{\mathrm dx}[x^2+y^2+8]}{x^2+y^2+8}=0$

We have by the chain rule,

$\dfrac{\mathrm d}{\mathrm dx}[x^2+y^2+8]=2x+2y\dfrac{\mathrm dy}{\mathrm dx}$

so that

$\dfrac{\mathrm dy}{\mathrm dx}\left(\ln(x^2+y^2+8)+\dfrac{2y^2}{x^2+y^2+8}\right)=-\dfrac{2xy}{x^2+y^2+8}$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2xy}{(x^2+y^2+8)\left(\ln(x^2+y^2+8)+\frac{2y^2}{x^2+y^2+8}\right)}$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2xy}{(x^2+y^2+8)\ln(x^2+y^2+8)+2y^2}$