I have this equation $x^3-xy^2+y^3=0$ and I want to know the value of the derivative at $(0,0)$. Through implicit differentiation I find $y'=\frac{y^2-3x^2}{3y^2-2xy}$. Now for $x=0,y=0$ this fraction becomes an indeterminate form. Upon graphical inspection I think if I were to draw a tangent at the Origin, it will have a slope very close to $-0.75$, but what is an easy method to find out the actual value? Note, when you graph the original function it LOOKS like a line, but it ISN'T. Thank you for your input.
Implicit differentiation and taking limit on derivative
calculusmultivariable-calculus
Best Answer
Because the function is homogeneous (every term is has degree $3$), the graph is actually in fact a union of lines through the origin. Since the cubic polynomial $u^3-u^2+1=0$ has precisely one real root $u_0$, the curve is in fact just the line $y=u_0x$.
Here's a general approach to such problems, if you're interested. Substitute $y=ux$ and factor out the highest power possible of $x$. In this case, you're left with $x^3(u^3-u^2+1)=0$, so in the $ux$-plane we get $x=0$ and $u^3-u^2+1=0$. This tells us that our curve becomes (aside from $x=0$) the curve $u^3-u^2+1=0$, which we've already said is just the line $u=u_0$. This means that our equation reduces to $y=u_0x$, as we said.
Let me show you a different example. Suppose you had instead the curve $y^2-x^2-x^3=0$. If you try implicit differentiation at the origin, you have the same situation. Now if I substitute $y=ux$, I get $x^2(u^2-1+x) = 0$, and so I'm looking at the parabola $x=1-u^2$ in the $xu$-plane. When $x=0$, we get $u=1$ and $u=-1$. This tells me that the lines $y=x$ and $y=-x$ are the two tangent lines to our original curve at the origin.