In one line
If \(y\) is defined implicitly by \(f(x,y)=0\), then \(\dfrac{dy}{dx}=-\dfrac{f_x}{f_y}\), provided \(f_y\ne0\).
Explained simply
Some curves, like a circle \(x^2+y^2=25\), are not written as "\(y=\) something". You can still find the slope at any point. Partial derivatives give a one-line formula: the slope is minus (rate of change in \(x\)) divided by (rate of change in \(y\)). It is quicker and less error-prone than differentiating both sides term by term.
Technically, since \(f(x,y)=0\) for all points on the curve, its total differential is zero: \(f_xdx+f_ydy=0\). Dividing by \(dx\) gives \(\dfrac{dy}{dx}=-\dfrac{f_x}{f_y}\). This is a self-learning topic in the syllabus. The same idea extends to surfaces: if \(f(x,y,z)=0\) defines \(z\) as a function of \(x\) and \(y\), its partial derivatives follow the same "minus ratio" rule with \(f_z\) in the denominator. This is useful when an equation cannot be solved for \(z\) at all.
How it works
- Bring everything to one side so that \(f(x,y)=0\).
- Find \(f_x\) treating \(y\) as constant and \(f_y\) treating \(x\) as constant.
- Write \(\dfrac{dy}{dx}=-\dfrac{f_x}{f_y}\) and simplify.
- Substitute a point if the slope at that point is asked.
Key formula
\(\dfrac{dy}{dx}=-\dfrac{\partial f/\partial x}{\partial f/\partial y}\), valid where \(f_y\ne0\). For \(f(x,y,z)=0\) defining \(z\): \(\dfrac{\partial z}{\partial x}=-\dfrac{f_x}{f_z}\), \(\dfrac{\partial z}{\partial y}=-\dfrac{f_y}{f_z}\).
Quick example
Circle \(x^2+y^2-25=0\): \(f_x=2x\), \(f_y=2y\), so \(\dfrac{dy}{dx}=-\dfrac xy\). At \((3,4)\) the slope is \(-\dfrac34\), which is perpendicular to the radius slope \(\dfrac43\), as geometry says it should be.
Folium \(x^3+y^3-3axy=0\): \(f_x=3x^2-3ay\), \(f_y=3y^2-3ax\), so \(\dfrac{dy}{dx}=-\dfrac{x^2-ay}{y^2-ax}=\dfrac{ay-x^2}{y^2-ax}\).
Common mistakes
- Dropping the minus sign in \(-f_x/f_y\).
- Not moving all terms to one side first when variables appear on both sides, e.g. \(x^y=y^x\) must become \(f=x^y-y^x=0\).
- Writing the ratio upside down as \(-f_y/f_x\).
How MU asks it
As a self-learning topic it usually appears as a 3-mark part: "If \(x^y+y^x=c\), find \(\dfrac{dy}{dx}\)" (answer \(-\dfrac{yx^{y-1}+y^x\log y}{x^y\log x+xy^{x-1}}\)). Write \(f(x,y)\), both partials with the variable held constant mentioned, the formula, and the simplified answer.