Implicit Differentiation for Plane Curves

Why Differentiate an Equation That Is Not Solved for \(y\)?

A plane curve can be described by an equation relating \(x\) and \(y\), even when it is inconvenient or impossible to write \(y\) explicitly as a function of \(x\). For instance, the circle \(x^2+y^2=25\) is one curve, but solving for \(y\) produces two branches, \(y=\sqrt{25-x^2}\) and \(y=-\sqrt{25-x^2}\). Differentiating the original equation lets us find the slope at a point without choosing and differentiating a branch formula.

The key is to remember that, while moving along a curve, \(y\) may change as \(x\) changes. We therefore treat \(y\) as a function of \(x\) and use the chain rule on every term that depends on \(y\).

Definition: An implicit equation \(F(x,y)=0\) describes the points \((x,y)\) for which the function \(F\) equals zero. To find a slope by implicit differentiation, suppose the curve can be followed near the point as a differentiable graph \(y=y(x)\), substitute that dependence into the equation, and differentiate with respect to \(x\).

The Chain Rule Along an Implicit Curve

Write the equation along a differentiable branch as \(F(x,y(x))=0\). Since the left side is a composite function of \(x\), its derivative is zero. The multivariable chain rule says that the derivative of \(F(x,y(x))\) is the partial derivative with respect to \(x\), plus the partial derivative with respect to \(y\) multiplied by \(dy/dx\). Thus, at each point on that branch,

F_x(x,y)+F_y(x,y)\frac{dy}{dx}=0.

Here \(F_x\) means differentiate \(F\) with respect to \(x\) while holding \(y\) fixed, and \(F_y\) means differentiate with respect to \(y\) while holding \(x\) fixed. The appearance of \(dy/dx\) is precisely the chain rule: a change in \(x\) changes \(F\) directly and may also change it indirectly through \(y(x)\).

If \(F_y(x,y)\ne 0\) at the point, ordinary algebra isolates the slope:

\frac{dy}{dx}=-\frac{F_x(x,y)}{F_y(x,y)}.

This calculation gives the slope of a differentiable graph branch through the point. It does not, by itself, establish that a branch exists there. If \(F_y=0\), the displayed equation cannot be solved for a finite \(dy/dx\) by division; the differentiated relation must be examined without dividing.

Why the method works: Along the curve, \(F(x,y(x))\) has the constant value zero. Its rate of change with respect to \(x\) must therefore be zero. Applying the chain rule to that composition gives the equation from which the slope is found.

A Reliable Differentiation Procedure

Keep the variables and their roles clear: \(x\) is the independent variable, while \(y\) varies with \(x\). In particular, differentiating \(y^2\) gives \(2y(dy/dx)\), not merely \(2y\). The same principle applies when \(y\) occurs inside a product, a power, or a trigonometric function.

Check the point.
Substitute its coordinates into the original equation to confirm that it lies on the curve.

Differentiate with respect to \(x\).
Use the chain rule on every expression involving \(y\), writing \(dy/dx\) explicitly.

Substitute the point.
Insert the known \(x\) and \(y\) coordinates into the differentiated equation.

Solve and check.
Find \(dy/dx\) if the equation allows it, then substitute the result back into the differentiated equation.

Worked Examples

Worked Example: Slope on a Circle

Find the slope of \(x^2+y^2=25\) at \((3,4)\). First check the point: \(3^2+4^2=9+16=25\), so it lies on the circle.

Differentiate both sides with respect to \(x\). The \(y^2\) term requires the chain rule because \(y\) depends on \(x\):

\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(25), \qquad 2x+2y\frac{dy}{dx}=0.

Substitute \(x=3\) and \(y=4\):

2(3)+2(4)\frac{dy}{dx}=0, \qquad 6+8\frac{dy}{dx}=0, \qquad \frac{dy}{dx}=-\frac{6}{8}=-\frac{3}{4}.

The slope is \(-3/4\). Check it in the point-specific differentiated equation: \(6+8(-3/4)=6-6=0\). The original equation was also satisfied by the point, as checked above.

Worked Example: A Polynomial Curve with a Product Term

Find the slope of \(x^2+xy+y^2=7\) at \((1,2)\). The point check gives \(1^2+(1)(2)+2^2=1+2+4=7\).

Differentiate term by term. For the product \(xy\), use the product rule, remembering that \(y\) varies with \(x\):

2x+\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}=0.

Group the terms that contain \(dy/dx\):

2x+y+(x+2y)\frac{dy}{dx}=0.

At \((1,2)\), this becomes

2(1)+2+\bigl(1+2(2)\bigr)\frac{dy}{dx}=0, \qquad 4+5\frac{dy}{dx}=0, \qquad \frac{dy}{dx}=-\frac{4}{5}.

Check the result in the point-specific equation: \(4+5(-4/5)=4-4=0\). Thus the slope at \((1,2)\) is \(-4/5\).

Worked Example: A Trigonometric Curve

Find the slope of \(\sin x+\cos y=0\) at \((0,\pi/2)\). The point lies on the curve because \(\sin 0+\cos(\pi/2)=0+0=0\).

Differentiate with respect to \(x\). The derivative of \(\cos y\) includes both the derivative of cosine and the chain rule factor \(dy/dx\):

\cos x-\sin y\frac{dy}{dx}=0.

Substitute \(x=0\) and \(y=\pi/2\):

\cos 0-\sin(\pi/2)\frac{dy}{dx}=0, \qquad 1-\frac{dy}{dx}=0, \qquad \frac{dy}{dx}=1.

Substitution checks the result: \(1-\sin(\pi/2)(1)=1-1=0\). The slope at the specified point is \(1\).

When the Differentiated Equation Does Not Give a Finite Slope

The differentiated relation is \(F_x+F_y(dy/dx)=0\). If \(F_y=0\) at the point but \(F_x\ne 0\), no finite value of \(dy/dx\) can satisfy that relation. This warns that the curve cannot be represented there by a differentiable graph \(y=y(x)\) with finite slope. In a regular plane curve, such a point may have a vertical tangent. If both partial derivatives are zero, the relation becomes \(0=0\) and does not determine the slope; further analysis is needed.

These cases are why checking the coefficient of \(dy/dx\) matters. Dividing by zero is never a valid way to obtain a slope, and the differentiated equation should be retained when its coefficient vanishes.

Common Mistake: Treating \(y\) as a Constant

A frequent error is to differentiate \(y^2\) as \(2y\), as though \(y\) did not change with \(x\). For the circle example, that incorrect step would give \(2x+2y=0\). At \((3,4)\), it says \(6+8=14=0\), which is false and cannot describe the slope.

The correction is to recognize \(y=y(x)\) and apply the chain rule: \(\frac{d}{dx}(y^2)=2y(dy/dx)\). The proper equation at \((3,4)\) is \(6+8(dy/dx)=0\), which gives \(-3/4\); substitution verifies \(6+8(-3/4)=0\). Whenever a term contains \(y\), ask whether differentiating it requires a factor of \(dy/dx\).

Why Implicit Differentiation Is Useful

Some curves are naturally given by a relation rather than a single explicit formula. Implicit differentiation avoids solving that relation first and works locally at a point, including when the equation represents several branches. It is also a direct application of the multivariable chain rule: the entire calculation follows from differentiating a constant value along a curve.

For reliable results, check the point in the original equation, include the chain rule factor whenever \(y\) varies, and verify the calculated slope in the differentiated equation. If the coefficient of the slope is zero, do not divide; interpret what the equation does and does not determine.

Check Your Understanding

For each question, make the dependence of \(y\) on \(x\) explicit when differentiating.

For \(x^2+y^2=10\), find the slope at \((1,3)\) and check it in the differentiated equation.
Differentiate \(x^2+xy=6\) with respect to \(x\), keeping the product rule terms visible.
For \(\sin x+\cos y=0\), write the differentiated relation before substituting a point.
Why is \(\frac{d}{dx}(y^3)=3y^2(dy/dx)\), rather than just \(3y^2\)?
What does the differentiated relation tell you if the coefficient of \(dy/dx\) is zero while the remaining term is nonzero?