Why the Point Must Be Checked First
An implicit equation describes a relationship between \(x\) and \(y\), rather than giving \(y\) as a formula in \(x\). Once implicit differentiation has produced a slope formula, evaluating it at a specified point may look like a simple substitution. But substitution is only the final step.
First, the point must actually satisfy the original relation. Next, the expressions in the equation and its derivative must be defined there. Finally, the conditions for the slope formula being used must hold. Skipping these checks can produce a numerical answer that is not the slope of the curve at the requested point.
The Implicit Slope Formula and Its Conditions
Write an implicit relation in the form \(F(x,y)=0\), where \(F\) is continuously differentiable in a neighborhood of a point \((a,b)\). The point must satisfy \(F(a,b)=0\). If \(F_y(a,b)\ne 0\), the implicit function theorem guarantees that near \((a,b)\), the relation can be represented by a differentiable function \(y=g(x)\), with \(g(a)=b\).
Here \(F_x\) means the partial derivative with respect to \(x\), holding \(y\) fixed, and \(F_y\) means the partial derivative with respect to \(y\), holding \(x\) fixed. When differentiating \(F(x,g(x))=0\) with respect to \(x\), the chain rule gives \(F_x+F_y g'(x)=0\). At \(x=a\), division by \(F_y(a,b)\) is allowed precisely because the theorem requires it to be nonzero.
A Reliable Evaluation Process
Use this process whenever a problem asks for the implicit slope at a specified point. The checks are part of the calculation, not optional formalities.
Substitute both coordinates into \(F(x,y)=0\). Confirm that the resulting equality is true.
Make sure the original equation and its derivatives are defined near the point. For example, a square root or logarithm may impose restrictions.
Use implicit differentiation and solve for \(dy/dx\), keeping track of terms that contain it.
For the theorem above, verify \(F_y(a,b)\ne 0\) before dividing by it.
Evaluate the formula at the point, then substitute the resulting slope into the differentiated equation at that point.
Worked Examples
Worked Example: Slope on a Circle
Find the implicit slope at \((3,4)\) on the circle \(x^2+y^2=25\).
Step 1: Check membership. Substitute the coordinates into the original equation:
Thus \((3,4)\) lies on the circle.
Step 2: Differentiate. Differentiating both sides with respect to \(x\), with \(y\) depending on \(x\), gives
Solving for the derivative:
Step 3: Check the denominator and evaluate. At the point, \(y=4\ne 0\), so the formula is defined. Therefore,
Check: Substitute \(x=3\), \(y=4\), and \(dy/dx=-3/4\) into the differentiated equation:
The derivative equation is satisfied, confirming the computed slope.
Worked Example: A Polynomial Relation with a Product
Find the implicit slope at \((1,2)\) on \(x^2+xy+y^2=7\).
Step 1: Check membership.
The point lies on the relation.
Step 2: Differentiate. Apply the product rule to \(xy\), remembering that \(y\) depends on \(x\):
Collect the derivative terms and solve:
Step 3: Check the denominator and substitute. At \((1,2)\), the denominator is \(1+2(2)=5\), which is nonzero. The slope is
Check: Substitute the point and slope into the differentiated equation:
The differentiated equation is satisfied, confirming the slope \(-\tfrac45\).
Worked Example: A Relation with Square Roots
Find the implicit slope at \((4,1)\) on \(\sqrt{x}+\sqrt{y}=3\).
Step 1: Check membership and domain. Both coordinates are positive, so the square roots are defined and differentiable near the point. Also,
so \((4,1)\) satisfies the relation.
Step 2: Differentiate. The chain rule gives
Solving for the derivative:
Step 3: Check the division. The coefficient \(1/(2\sqrt{y})\) is defined and nonzero at \(y=1\). Substitute the point:
Check:
The differentiated relation is satisfied at the point.
Why the Conditions Matter
There are two distinct questions: does the point lie on the relation, and can the implicit-function formula be used there? Both must be addressed before reporting a slope from that formula.
For example, the expression \(-x/y\) came from differentiating \(x^2+y^2=25\) and solving for \(dy/dx\). At \((1,2)\), this expression has the numerical value \(-1/2\), but
So \((1,2)\) is not on the circle. The number \(-1/2\) is not the slope of that circle at the specified point. The first check prevents an algebraically computable expression from being mistaken for a slope on the requested curve.
The denominator check has a different purpose. If \(F_y(a,b)=0\), the formula \(-F_x/F_y\) cannot be evaluated there, and the implicit-function theorem in the stated form does not justify using it. This does not prove that no slope exists. For instance, \(F(x,y)=(y-x)^2\) describes the line \(y=x\), whose slope is \(1\), while \(F_y=2(y-x)=0\) on that line. In such a case, a slope, if claimed, must be established by another argument rather than by dividing by \(F_y\).
Common Mistake: Substituting Before Checking
A common error is to differentiate an equation, substitute the requested coordinates into the resulting expression, and stop as soon as a number appears. That procedure skips the test that the point satisfies the original equation. A second error is to divide by a coefficient without checking that it is nonzero at the point. Either can make an invalid calculation look complete.
Check Your Understanding
For each question, make the membership and formula checks explicit before giving a slope.
- For \(x^2+y^2=13\), check whether \((2,3)\) lies on the relation. If it does, find the implicit slope there.
- For \(xy+y^2=6\), find the slope at \((2,2)\), verifying the point and the denominator in your formula.
- For \(\sqrt{x}+\sqrt{y}=5\), explain whether \((9,4)\) is in the domain and on the relation, then evaluate the implicit slope if the formula applies.
- Why does \(F_y(a,b)=0\) prevent use of the formula \(-F_x(a,b)/F_y(a,b)\)? Does it by itself prove that no slope exists?
- A student evaluates a derivative formula at a pair that does not satisfy the original equation. What is wrong with treating the result as the slope at that pair?
