Differentiating Both Sides of an Implicit Equation

Why Differentiate Both Sides?

An equation involving \(x\) and \(y\) can describe a relationship without writing \(y\) explicitly as a formula in terms of \(x\). For example, \(x^2+y^2=25\) describes points on a circle. At a part of that curve where \(y\) varies differentiably with \(x\), we can still find the rate \(\frac{dy}{dx}\) without first solving the equation for \(y\).

The essential idea is to regard \(y\) as a function of \(x\) while differentiating. This means a term involving \(y\) is a composite function, so its derivative includes the Chain Rule factor \(\frac{dy}{dx}\). Then we differentiate each side of the equation and rearrange to isolate that factor.

The Basic Procedure

Suppose an equation remains true as \(x\) varies along a curve and \(y=y(x)\) is differentiable on the portion under consideration. Differentiate the left side and the right side with respect to \(x\). The derivative of a term involving \(x\) uses the usual differentiation rules. The derivative of a term involving \(y\) also uses those rules, but the Chain Rule accounts for the dependence of \(y\) on \(x\).

For instance, the derivative of \(y^2\) is not just \(2y\). Since \(y\) depends on \(x\), the Chain Rule gives \(2y\frac{dy}{dx}\). Similarly, the derivative of \(y^3\) is \(3y^2\frac{dy}{dx}\). By contrast, the derivative of \(x^2\) with respect to \(x\) is simply \(2x\).

Worked Example: Differentiating a Circle Equation

Why the Chain Rule Factor Appears

The factor \(\frac{dy}{dx}\) is not an extra convention added for implicit equations. It follows directly from the Chain Rule. If \(y=y(x)\), then \(y^2\) means the composite expression \((y(x))^2\). Differentiating this composite expression gives the derivative of the outer squaring function, multiplied by the derivative of the inner function.

This reasoning works for other differentiable functions of \(y\) as well. For example, the Chain Rule gives \(\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}\). In every case, the derivative of the inner quantity \(y\) with respect to \(x\) must be included.

A General Justification

The method can be justified by treating the equation as an equality of functions along a differentiable part of a curve. If the equation is \(L(x,y)=R(x,y)\), then after writing \(y=y(x)\), both sides are functions of \(x\). Since equal differentiable functions have equal derivatives, differentiating each side gives another equality. Every occurrence of \(y\) is a composite function of \(x\), so the Chain Rule supplies its factor of \(\frac{dy}{dx}\).

Proof. Define two functions of \(x\) by \(u(x)=L(x,y(x))\) and \(v(x)=R(x,y(x))\). The assumed relation says \(u(x)=v(x)\) throughout the interval. Differentiability then implies \(u'(x)=v'(x)\) at every point where these derivatives exist, because the difference \(u-v\) is the constant zero function and therefore has derivative zero. When calculating \(u'\) or \(v'\), any expression depending on \(y(x)\) is a composite function, so its derivative follows the Chain Rule. Thus differentiating the two sides of the original relation produces an equality that is valid along the curve. Rearranging that equality isolates \(\frac{dy}{dx}\) whenever its coefficient is nonzero. This proves the procedure.

Worked Example: Terms in \(y\) on Both Sides

When Can You Solve for the Derivative?

After differentiating and rearranging, the equation often has a term multiplied by \(\frac{dy}{dx}\), along with terms that do not contain the derivative. If that coefficient is nonzero, division isolates \(\frac{dy}{dx}\). The condition matters: division by zero is not valid, so the resulting formula applies only where the coefficient is nonzero.

This procedure finds a derivative provided \(y\) is differentiable as a function of \(x\) on the portion being considered and the differentiated equation can be solved for \(\frac{dy}{dx}\). If the coefficient of \(\frac{dy}{dx}\) is zero, the algebra does not produce a formula by division at that point. Do not silently divide by that coefficient or assume the same formula applies there.

Worked Example: A Cubic Term in \(y\)

Common Mistake: Forgetting That \(y\) Depends on \(x\)

A frequent error is to differentiate \(y^2\) as though \(y\) were an independent constant with respect to \(x\). That misses the fact that \(y\) changes as \(x\) changes.

Incorrect calculation: From \(x^2+y^2=25\), someone might write

This is wrong because the derivative of \(y^2\) is not \(2y\) when \(y\) depends on \(x\). The missing inner derivative is \(\frac{dy}{dx}\).

Correction: Apply the Chain Rule to the \(y^2\) term:

The correctly differentiated equation is therefore

which can be solved for \(\frac{dy}{dx}\) when \(y\ne 0\).

A Reliable Routine

For a basic implicit differentiation problem, keep the following sequence in mind:

1. Write the original relation clearly and identify both sides.
2. Differentiate each side with respect to \(x\).
3. For every term involving \(y\), apply the Chain Rule and include \(\frac{dy}{dx}\).
4. Rearrange the differentiated equation to isolate \(\frac{dy}{dx}\).
5. State any condition required for division, then check the result in the differentiated equation.

The key habit is to attach the Chain Rule factor at the moment you differentiate a \(y\)-term. Once that is done consistently, differentiating both sides turns an implicit relation into an equation that can often be solved for the derivative.