Implicit Differentiation

The central idea

An equation $F(x,y)=0$ may define $y$ locally as a differentiable function of $x$ when $F$ is differentiable and $F_y\ne0$. Differentiating along the curve gives $F_x+F_y y'=0$, hence $y'=-F_x/F_y$. Every derivative of a $y$-expression must include $dy/dx$ by the chain rule.

Definitions, hypotheses, and notation

The formula $y'=-F_x/F_y$ exposes a boundary case: where $F_y=0$, solving locally for $y$ as a differentiable function of $x$ may fail. A curve may instead have a vertical tangent, and it may be possible to solve for $x$ as a function of $y$ if $F_x\ne0$. The algebraic denominator therefore has geometric meaning.

Second derivatives require another round of implicit differentiation and product rules because $y'$ also depends on $x$. It is usually clearest to isolate $y'$ first, differentiate that relation, and only then substitute a point. Even for first derivatives, substitution after differentiation preserves which quantities vary along the curve.

Conceptual meaning

Implicit differentiation finds tangent slopes without first solving for $y$. The gradient $\nabla F=\langle F_x,F_y\rangle$ is normal to the level curve, while the tangent direction $\langle1,y'\rangle$ is perpendicular to it.

A dependable method and decision rule

1. Differentiate both sides with respect to $x$.
2. Treat $y$ as the composite function $y(x)$.
3. Attach a factor $y'$ whenever differentiating a function of $y$.
4. Collect all terms containing $y'$ on one side and factor it out.
5. Solve for $y'$ before substituting the coordinates of a requested point.

Fully worked example

Graphical or geometric meaning

At $(2,2)$ the gradient is $\langle4,16\rangle$, normal to the ellipse. A tangent direction is $\langle1,-1/4\rangle$; their dot product is $4-4=0$, confirming perpendicularity.

Common mistakes and why they fail

Verification and reasonableness checks

• Substitute the point into the original relation.
• Verify the tangent direction is perpendicular to the gradient.
• If solving explicitly is easy, differentiate one branch and compare.

Track every dependent quantity

An equation relating $x$ and $y$ treats $y$ as $y(x)$, so differentiating $y^n$ gives $ny^{n-1}y'$. Differentiate every term, collect all factors of $y'$, and only then solve. A zero denominator in the resulting formula may indicate a vertical tangent, but check the numerator and original equation first. A point not on the curve cannot be used. For a circle, geometry supplies a check because tangent and radius are perpendicular. If numerator and denominator both vanish, the simple formula is inconclusive; the curve may have several branches or a singular point, and local analysis is needed instead of automatic cancellation.

Practice

1. For $x^2+y^2=25$, find $y'$.
2. Find the slope on that circle at $(3,4)$.
3. Differentiate $xy=6$ implicitly.

Connections and next steps

• Chain Rule
• Derivative
• Tangent Planes