Implicit Differentiation

3. Visual Connection

5. Concept Development

An equation can describe a curve without isolating \(y\). For example, \(x^2+y^2=25\) represents an entire circle, even though the circle is not the graph of one function of \(x\). Near most points, however, a small branch of the curve behaves like a differentiable function \(y=y(x)\). Implicit differentiation finds the slope of that local branch directly from the equation.

The central idea is the chain rule. Because \(y\) depends on \(x\), every time a derivative passes through an expression involving \(y\), it produces a factor of \(dy/dx\):

By contrast, \(d(x^n)/dx=nx^{n-1}\) has no extra factor because \(dx/dx=1\).

Reliable procedure.

1. Verify that the point, if one is supplied, satisfies the original equation.
2. Differentiate every term on both sides with respect to \(x\), treating \(y\) as \(y(x)\).
3. Move all terms containing \(y'\) to one side and all remaining terms to the other.
4. Factor out \(y'\), then divide to solve for \(dy/dx\).
5. Substitute the point only after deriving the slope formula, unless point substitution clearly reduces the algebra.

Compact structural form. If a curve is written as \(F(x,y)=C\), differentiating gives

whenever \(F_y\ne0\). This formula summarizes the term-by-term method; it does not replace the need to differentiate accurately.

Reading special tangents. After simplifying \(y'=N/D\), a horizontal tangent is a candidate where \(N=0\) and \(D\ne0\). A vertical tangent is a candidate where \(D=0\) and \(N\ne0\). Every candidate must also lie on the original curve. If both numerator and denominator are zero, the formula alone is inconclusive.

Local branches matter. An implicit equation may describe several \(y\)-values for one \(x\). The derivative is attached to a particular point and branch, so two points with the same \(x\)-coordinate can have different slopes. A zero denominator does not mean the curve disappears; it may signal a vertical tangent or a point requiring further analysis.

6. Detailed Worked Example and Error Check

Example 1: Polynomial relation and tangent line. Consider

Differentiate term by term. The middle term requires the product rule:

Collect and factor the \(y'\)-terms:

The point \((1,2)\) lies on the curve because \(1+2+4=7\). Its slope is \(-4/5\), so the tangent line is \(y-2=-\frac45(x-1)\).

Example 2: A trigonometric term containing \(y\). For

the chain rule differentiates the input \(x+y\):

Solving for the derivative gives

Example 3: Products and nested powers. Differentiate \((x+y)^3=x^2y\):

Separating the derivative terms produces

Example 4: Horizontal and vertical tangents. On the ellipse \(4x^2+9y^2=36\),

Horizontal tangents require \(x=0\) and \(y\ne0\). The curve equation gives \((0,2)\) and \((0,-2)\). Vertical tangents require \(y=0\) and \(x\ne0\), giving \((3,0)\) and \((-3,0)\).

Example 5: Find a second derivative. From the circle \(x^2+y^2=25\),

Differentiate this first-derivative equation again:

Substituting \(y'=-x/y\) and using \(x^2+y^2=25\) gives

Notice that differentiating \(yy'\) requires the product rule: its derivative is \((y')^2+yy''\).

Example 6: An inconclusive derivative equation. Consider \(x^3+y^3=3xy\) at \((0,0)\). Differentiation gives

At \((0,0)\), the equation reduces to \(0=0\), which gives no unique slope. This is a warning: when both derivative coefficients cancel, the point needs separate local analysis. Mechanical division would hide the indeterminate structure.

AP error check. Do not omit \(y'\) after differentiating a \(y\)-expression, forget the product rule for \(xy\), substitute a point that is not on the curve, cancel a factor that may be zero at the target point, or identify tangent candidates without checking the original equation.