Why Products and Quotients Need Care
Implicit equations often combine \(x\) and \(y\) in the same product or quotient. For example, a relation may contain \(xy\), \(x^2y\), or \(y/x\). When differentiating such an equation with respect to \(x\), \(y\) is not a constant: along a differentiable branch of the curve, \(y\) depends on \(x\).
That means a product such as \(xy\) requires the product rule, and a quotient such as \(y/x\) requires the quotient rule. Any derivative of an expression involving \(y\) must also account for the chain rule. After differentiating, the goal is to collect the terms containing \(dy/dx\) and solve for that derivative.
Rules for Implicit Products and Quotients
If \(u=u(x)\) and \(v=v(x)\) are differentiable, their product and quotient rules are
In an implicit equation, \(u\) and \(v\) may themselves depend on \(y(x)\). For instance, the derivative of \(xy\) is not just \(y\): both factors vary, so
Likewise, the derivative of \(y^2\) is \(2y\,dy/dx\), not \(2y\), because \(y\) is a function of \(x\). It is useful to write \(y'\) for \(dy/dx\) during the calculation, then use either notation consistently.
Why the Quotient Rule Has Its Form
The quotient rule follows from the product rule and the chain rule. If \(v(x)\ne 0\), write \(u/v=u(v^{-1})\). The derivative of \(v^{-1}\), by the chain rule applied to the power \(v^{-1}\), is \(-v^{-2}v'\). Therefore,
This also explains the order of the terms in the numerator: it is derivative of the top times the bottom, minus the top times derivative of the bottom. The denominator is squared.
A Reliable Differentiation Process
For an equation involving products or quotients, keep the original structure visible until the relevant rule has been applied. Then gather every term containing \(dy/dx\) on one side. The sequence below helps prevent missing product terms or sign errors.
Mark products and quotients, and check that any denominator is nonzero on the part of the relation being considered.
Apply the product or quotient rule to each complete expression. When differentiating \(y\) or a function of \(y\), include the factor \(dy/dx\) required by the chain rule.
Move all terms containing \(dy/dx\) to one side and all other terms to the other side, preserving signs.
Factor out \(dy/dx\), divide only when the coefficient is nonzero, and substitute the result into the differentiated equation to verify it.
Worked Examples
Worked Example: A Product and a Power of \(y\)
Find \(dy/dx\) along a differentiable branch of \(xy+y^2=x^2+3\). Then evaluate the derivative at \((1,1)\).
Differentiate each side with respect to \(x\). The product \(xy\) requires the product rule, \(y^2\) requires the chain rule, and \(x^2+3\) differentiates to \(2x\):
Collect the terms containing \(dy/dx\):
Where \(x+2y\ne 0\), divide by that coefficient:
First check that the point lies on the original relation: \(1\cdot1+1^2=2\), while \(1^2+3=4\). This shows \((1,1)\) is not on this relation, so it cannot be used to evaluate its implicit derivative. Instead, use the point \((1,2)\): the left side is \(1\cdot2+2^2=6\), and the right side is \(1+3=4\), so that point is also not on the relation. Thus this equation has no point \((1,1)\) or \((1,2)\) suitable for the proposed evaluation.
To evaluate at a valid point, use \((2,1)\): the left side is \(2\cdot1+1=3\), and the right side is \(4+3=7\), so it is not on the relation either. The relation has no convenient integer point of this form; the differentiation formula itself is the result, and it applies at points on the curve where \(x+2y\ne0\).
Worked Example: Two Product Terms
Find the implicit derivative for \(x^2y+xy^2=6\), and evaluate it at the point \((1,2)\).
Check the point first: \(1^2\cdot2+1\cdot2^2=2+4=6\), so it lies on the relation. Differentiate both product terms. For \(x^2y\), the product rule gives \(2xy+x^2\,dy/dx\). For \(xy^2\), it gives \(y^2+2xy\,dy/dx\).
Collect the derivative terms and the remaining terms:
At \((1,2)\), the coefficient of \(dy/dx\) is \(1^2+2(1)(2)=5\), and the right side is \(-(2(1)(2)+2^2)=-8\). Therefore,
Check by substituting \(x=1\), \(y=2\), and \(dy/dx=-8/5\) into the differentiated equation:
The differentiated equation is satisfied, confirming the calculated derivative.
Worked Example: A Quotient with \(y\) in the Denominator
For \(y/x+x/y=5/2\), find \(dy/dx\) at \((1,2)\). The quotients require \(x\ne0\) and \(y\ne0\), both of which hold at this point.
Check the relation: \(2/1+1/2=2+1/2=5/2\). Differentiate each quotient using the quotient rule:
The right side \(5/2\) is constant, so its derivative is zero. Thus
Substitute \(x=1\) and \(y=2\):
Multiply each term by \(4\):
Therefore \(3\,dy/dx-6=0\), giving \(dy/dx=2\). Check in the differentiated equation: the first quotient derivative is \((1\cdot2-2)/1^2=0\); the second is \((2-1\cdot2)/2^2=0\). Their sum is zero, as required.
Worked Example: A Quotient of Product Expressions
Find the derivative at \((2,2)\) for the relation \(\frac{xy}{x+y}=1\). The denominator \(x+y\) is \(4\) at the point, so the quotient is defined there.
Check the original equation: \((2\cdot2)/(2+2)=4/4=1\). Use the quotient rule with numerator \(xy\) and denominator \(x+y\). Their derivatives are \(y+x\,dy/dx\) and \(1+dy/dx\), respectively:
At \((2,2)\), this becomes
Expand the numerator and simplify:
So \(4+4\,dy/dx=0\), and \(dy/dx=-1\). Check by substituting \(-1\) into the numerator: \((2+2(-1))4-4(1+(-1))=0-0=0\). The quotient derivative is therefore zero, as the differentiated relation requires.
Common Mistake: Forgetting One Product Term
A frequent error is to treat \(y\) as though it were constant while differentiating a product. For \(xy+y^2=x^2+3\), an incorrect calculation might write
This omits the derivative of the factor \(x\) in \(xy\). The correct product rule is \(\frac{d}{dx}(xy)=y+x\,dy/dx\), so the correct differentiated equation is
The omitted \(y\) changes the equation and therefore changes the derivative. A quick safeguard is to label the two factors before applying the product rule: derivative of the first times the second, plus the first times derivative of the second.
What to Check Before Solving
After differentiating, the coefficient of \(dy/dx\) may be zero at a point. In that case, dividing by the coefficient is not valid, and the equation does not yield a slope by this algebraic step. Likewise, a quotient rule calculation is not valid where its denominator is zero. These are conditions on the calculation, not details to ignore after finding a formula.
For ordinary points where the required denominators are nonzero and the coefficient of \(dy/dx\) is also nonzero, the process is direct: differentiate each structured expression, collect derivative terms, solve, and check by substitution into the differentiated equation. The check is especially useful for catching a missing chain-rule factor or a sign error in the quotient numerator.
Check Your Understanding
For each question, show the product or quotient rule step before collecting terms.
- For \(xy+x^2=5\), differentiate implicitly and solve for \(dy/dx\) wherever the resulting coefficient allows division.
- Differentiate \(x^2y+3y^2=8\), making clear where the chain rule contributes \(dy/dx\).
- For \(y/x=3\), state the condition needed to use the quotient rule and find the resulting derivative relation.
- Differentiate \(xy/(x+y)=2\), and identify the condition on its denominator.
- Explain why differentiating \(xy\) as \(x\,dy/dx\) is incomplete.
