I was solving a problem on implicit differentiation and I was wondering why my answer does not match the given answer. Apparently I multiplied both parts of an equation by an expression to get rid of some fractions before differentiating. Here is my illustration:
Differentiating $$\frac{x}{\sqrt y}+\frac{y}{\sqrt x}=xy$$ we get $$\frac{\sqrt y-\frac{xy'}{2\sqrt y}}{y}+\frac{y'\sqrt x-\frac{y}{2\sqrt x}}{x}=y+xy'$$ or $$y'=\frac{y-\frac{1}{\sqrt y}+\frac{y}{2x\sqrt x}}{\frac{1}{\sqrt x}-x-\frac{x}{2y\sqrt y}}=\frac{2xy^2\sqrt{xy}-2xy\sqrt x+y^2\sqrt y}{2xy\sqrt y-2x^2y\sqrt {xy} -x^2\sqrt x}$$However if I multiply both parts by $\sqrt{ xy}$ then differentiate $x^{\frac{3}{2}}+y^{\frac{3}{2}}=x^{\frac{3}{2}}y^{\frac{3}{2}}$, the result will be: $$\frac{3}{2}x^{\frac{1}{2}}+\frac{3}{2}y^{\frac{1}{2}}y'=\frac{3}{2}x^{\frac{1}{2}}y^{\frac{3}{2}}+\frac{3}{2}x^{\frac{3}{2}}y^{\frac{1}{2}}y'$$ or
$$y'=\frac{y\sqrt{ xy}-\sqrt x}{\sqrt y-x\sqrt{ xy}}$$
Graphing both equations produces the same graph so the slope of a tangent line should be no different. Why do I get two different expressions for the slope?
Consider a simpler example. $$ f(x,y) = y-x=0,\qquad\text{implies}\qquad g(x,y)=y^2-xy = 0, $$
so from $f$ we find $y'(x)=1$; from $g$ we find $y'(x)=\frac{y}{2y-x}$, which gives the same answer if $y=x$ but is a different function of $x,y$ in general.
The equation $\frac{d}{dx}f(x,y(x))=0$ allows you to find $y'(x)$ not only on curve $f(x,y)=0$ but on the whole family of curves $f(x,y)=c$ (the same is true for $g$). When $c=0$ both of the families produce the same curve (*). However, in general, they produce different curves, so the slopes of those curves will be different as well.
(*) in the appropriate region