Why Check an Implicit Derivative?
An implicit derivative is often found by differentiating an equation in \(x\) and \(y\), then collecting the terms that contain \(y'\). The algebra can be compact, but a small sign error or an invalid division can change the answer. A reliable check asks whether the proposed derivative is consistent with the original relation and with the differentiated relation.
These checks answer different questions. Substituting a slope into the differentiated equation tests the derivative calculation. Substituting a point into the original equation tests whether that point belongs to the curve at all. Neither check should be mistaken for the other.
The Derivative Equation as a Check
Write a smooth implicit relation as \(F(x,y)=0\). If a curve on this relation can locally be described by a differentiable function \(y=y(x)\), then the chain rule gives
Here \(F_x\) and \(F_y\) are the partial derivatives of \(F\) with respect to its first and second variables. This equation is a necessary condition for the slope along the curve. If \(F_y\ne 0\), it can be solved for the slope:
This quotient must not be used at a point where \(F_y=0\). At such a point the differentiated relation may still be meaningful, but dividing by \(F_y\) is not. The relation may require a separate analysis to determine whether a derivative exists.
Why Substitution Verifies the Slope Equation
The check follows directly from the chain rule. Suppose a differentiable curve \(y=y(x)\) satisfies \(F(x,y(x))=0\). Since this expression is constantly zero along the curve, its derivative is zero. Therefore \(F_x+F_y y'=0\) at every point where the derivatives exist. A candidate slope passes this check precisely when substituting it makes that equation true.
This reasoning also shows why checking only the derivative equation is not enough. Differentiating a relation can discard information about which points satisfy the original equation. For instance, an algebraic manipulation might produce a point that satisfies a transformed or differentiated equation but not the starting relation. Always return to the original equation.
Worked Example: Verify a Slope on a Circle
Worked Example: Verify a Slope on a Circle
For the circle \(x^2+y^2=25\), check the implicit derivative at the point \((3,4)\).
First verify the point in the original relation:
So \((3,4)\) lies on the circle. Differentiate the original equation with respect to \(x\), treating \(y\) as a function of \(x\):
At \((3,4)\), this becomes
Now check the result in the differentiated relation:
The proposed slope satisfies the derivative equation, and the point satisfies the original equation. The check confirms the slope \(y'=-\frac{3}{4}\) at this point.
Compare with an Independent Calculation
A particularly strong check is to obtain the slope by a different route. If the relation can be solved for a local branch \(y\) as a function of \(x\), differentiate that expression and compare the answer with the implicit formula. Agreement between two calculations makes a sign or algebra error less likely.
Worked Example: Check an Implicit Derivative Against an Explicit Branch
Worked Example: Check an Implicit Derivative Against an Explicit Branch
Consider \(xy+y^2=6\), and check the slope at \((1,2)\) by two methods.
First, verify the point:
Differentiate the relation implicitly. The product rule gives \(\frac{d}{dx}(xy)=y+xy'\), and \(\frac{d}{dx}(y^2)=2yy'\). Therefore
Collect the terms containing \(y'\), then solve where \(x+2y\ne 0\):
At \((1,2)\), the denominator is \(1+2(2)=5\), so
Now solve the original relation for \(y\). The quadratic formula applied to \(y^2+xy-6=0\) gives
The branch passing through \((1,2)\) uses the plus sign, because \(\frac{-1+\sqrt{25}}{2}=2\). On this branch,
Differentiate it using the chain rule:
At \(x=1\), \(\sqrt{x^2+24}=\sqrt{25}=5\), so
The independent calculation agrees with the implicit derivative. As a final check, substitute the slope into the differentiated relation at \((1,2)\):
Both the original relation and its derivative equation are satisfied at the point with the computed slope.
Check the Original Relation After Algebraic Manipulation
Rewriting an equation can change which points it describes unless every step preserves the original conditions. Squaring both sides is a common source of extra solutions: if \(A=B\), then \(A^2=B^2\), but \(A^2=B^2\) also allows \(A=-B\). A point found from the squared equation must therefore be tested in the equation before squaring.
Worked Example: Reject a Point Introduced by Squaring
Worked Example: Reject a Point Introduced by Squaring
Consider the relation \(\sqrt{y+1}=x-1\). Its real-valued domain requires \(y\ge -1\), and the equality also requires \(x-1\ge 0\). Squaring gives
The point \((0,0)\) satisfies this squared equation, since \((0-1)^2-1=0\). But test it in the original relation:
Thus \((0,0)\) is not on the original relation. Any slope attributed to that point using the squared equation is irrelevant to the original curve.
For contrast, \((2,0)\) does satisfy the original relation because \(\sqrt{0+1}=1=2-1\). Differentiate the original equation at points where \(y+1>0\):
At \((2,0)\), this gives \(y'=2\). The squared equation gives \(y=(x-1)^2-1\), hence \(y'=2(x-1)\), which is also \(2\) at \(x=2\). The agreement at this valid point confirms the slope there; it does not make the extra point \((0,0)\) valid.
A Common Mistake: Checking Only the Differentiated Equation
A tempting shortcut is to take a point that solves a differentiated or transformed equation and assume that it belongs to the original curve. The example above shows why this fails: the squared equation accepts \((0,0)\), but the original equation rejects it. A derivative calculation cannot repair a point that was never on the relation.
Another error is dividing by a factor without checking whether it can be zero. For example, from \((x+2y)y'=-y\), the quotient \(y'=-y/(x+2y)\) is valid only when \(x+2y\ne0\). If \(x+2y=0\), division is not allowed; return to the undivided equation and test it at the point instead.
A Reliable Verification Procedure
Use the following sequence when checking an implicit derivative or a slope at a specified point:
Substitute its coordinates into the original relation, including any domain restrictions.
Apply the chain rule to every term that depends on \(y\), and retain the equation before dividing.
Substitute the point and the slope into the differentiated equation and simplify to zero.
Confirm that no division used a zero denominator and that transformations did not add points.
Compare with a derivative from an explicit branch or another direct calculation.
These checks are useful because an implicit derivative is not merely a formula to memorize. It is a slope tied to a particular point, the original relation, and the conditions under which the algebra is valid.
Check Your Understanding
For each question, distinguish the original relation from the equation obtained by differentiating it.
- For \(x^2+y^2=10\), verify the point \((1,3)\) and check the implicit slope at that point in the differentiated equation.
- Why does substituting a candidate slope into \(F_x+F_y y'=0\) not, by itself, prove that the chosen point lies on \(F(x,y)=0\)?
- For \(xy+y^2=6\), at \((1,2)\), compare the implicit slope with the derivative obtained from the positive quadratic-formula branch.
- What condition must hold before dividing by \(F_y\) to write \(y'=-F_x/F_y\)?
- Explain why a point satisfying an equation obtained by squaring must still be checked in the original equation.
