AcademyDifferential Calculus
Academy
Implicit Differentiation
Level 1 - Calculus topic page in Differential Calculus.
Principle
Implicit differentiation handles equations where \(y\) is defined by a relationship with \(x\), rather than explicitly as \(y=f(x)\). Treat \(y\) as a function of \(x\).
Notation
\(y\)
dependent variable, treated as y(x)
\(\frac{dy}{dx}\)
derivative of y with respect to x
\(F(x,y)=0\)
implicit relation between x and y
The Core Method
Differentiate both sides with respect to \(x\). Every time a term involving \(y\) is differentiated, multiply by \(dy/dx\).
Implicit chain rule
\[\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}\]
After differentiating, collect all \(dy/dx\) terms and solve for \(dy/dx\).
Worked Cases
Question
Find \(dy/dx\) if \(x^2+y^2=25\).
Answer
Differentiate both sides: \(2x+2y\frac{dy}{dx}=0\). Move \(2x\) to the other side: \(2y\frac{dy}{dx}=-2x\). Divide by \(2y\): \(\frac{dy}{dx}=-\frac{x}{y}\), where \(y\ne0\).
Examples
Question
Differentiate \(xy=1\) implicitly.
Answer
Use the product rule on \(xy\): \(\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\). The derivative of \(1\) is \(0\). Thus \(x\frac{dy}{dx}+y=0\), so \(\frac{dy}{dx}=-y/x\) when \(x\ne0\).