Why a Formula for the Slope Is Useful
Some plane curves are most naturally described by an equation relating \(x\) and \(y\), rather than by a formula \(y=f(x)\). A circle, for example, is described by \(x^2+y^2=25\). Near many points on that circle, \(y\) can be viewed as a function of \(x\), even though the equation does not initially give \(y\) explicitly.
The implicit derivative formula gives the slope of that local function directly from the defining equation. Its key idea is simple: if \(y\) depends on \(x\) along the curve, then differentiating \(F(x,y)=0\) requires the chain rule.
The Formula and Its Condition
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. The derivatives in the quotient are evaluated at the point \((x,y(x))\) on the curve.
The condition \(F_y(a,b)\ne 0\) matters in two ways. The implicit function theorem guarantees that the curve can locally be represented as a differentiable function \(y(x)\), and the quotient in the formula has a nonzero denominator. The theorem is local: it does not promise that one function \(y(x)\) describes the entire curve.
Deriving the Implicit Derivative Formula
Assume that the curve is represented near a point by a differentiable function \(y=y(x)\), and that \(F(x,y(x))=0\) throughout that interval. Differentiating this identity with respect to \(x\) gives the formula.
The equation becomes the identity \(F(x,y(x))=0\). Both the first input \(x\) and the second input \(y(x)\) vary as \(x\) changes.
The derivative of \(F(x,y(x))\) is \(F_x(x,y(x))+F_y(x,y(x))y'(x)\). The second term includes \(y'(x)\) because the second input depends on \(x\).
Since the constraint is identically zero, its derivative is zero: $$ F_x(x,y(x))+F_y(x,y(x))y'(x)=0. $$
Where \(F_y(x,y(x))\ne 0\), rearranging gives $$ y'(x)=-\frac{F_x(x,y(x))}{F_y(x,y(x))}. $$
This is the derivation: the formula is the chain rule applied to a constraint that stays equal to zero, followed by solving for \(y'(x)\). The implicit function theorem supplies the important existence statement: if \(F\) is continuously differentiable and \(F_y\ne0\) at the point, such a local differentiable function \(y(x)\) exists.
Worked Example: Slope on a Circle
Worked Example: Slope on a Circle
Find the slope of the circle \(x^2+y^2=25\) at \((3,4)\), using the implicit derivative formula.
Set \(F(x,y)=x^2+y^2-25\). Then $$ F_x(x,y)=2x,\qquad F_y(x,y)=2y. $$ The point lies on the curve because \(3^2+4^2-25=9+16-25=0\). Also, \(F_y(3,4)=2(4)=8\ne0\), so the formula applies.
Substitute the partial derivatives and the point: $$ \left.\frac{dy}{dx}\right|_{(3,4)} =-\frac{F_x(3,4)}{F_y(3,4)} =-\frac{2(3)}{2(4)} =-\frac{6}{8} =-\frac{3}{4}. $$
Check by differentiating the constraint along the curve: $$ 2x+2y\frac{dy}{dx}=0. $$ At \((3,4)\), this gives \(6+8(-3/4)=6-6=0\), as required.
Worked Example: A Curve with Mixed Terms
Worked Example: A Curve with Mixed Terms
Find the slope of \(x^2+xy+y^2=7\) at \((1,2)\).
Define \(F(x,y)=x^2+xy+y^2-7\). Check that the point satisfies the equation: $$ 1^2+(1)(2)+2^2-7=1+2+4-7=0. $$ The partial derivatives are $$ F_x(x,y)=2x+y,\qquad F_y(x,y)=x+2y. $$ At the point, \(F_x(1,2)=2(1)+2=4\), and \(F_y(1,2)=1+2(2)=5\ne0\). Therefore, $$ \left.\frac{dy}{dx}\right|_{(1,2)} =-\frac{4}{5}. $$
To check, differentiate each term along the curve. The derivative of \(x^2\) is \(2x\), the derivative of \(xy\) is \(y+xy'\) by the product rule, and the derivative of \(y^2\) is \(2yy'\). Thus $$ 2x+y+xy'+2yy'=0. $$ At \((1,2)\), substituting \(y'=-4/5\) gives $$ 2+2+1\left(-\frac45\right)+2(2)\left(-\frac45\right) =4-\frac45-\frac{16}{5} =4-\frac{20}{5} =0. $$ The differentiated constraint is satisfied.
Worked Example: A Nonlinear Dependence on \(y\)
Worked Example: A Nonlinear Dependence on \(y\)
For the curve \(y^3+xy-1=0\), find the slope at \((0,1)\).
Let \(F(x,y)=y^3+xy-1\). The point is on the curve because $$ 1^3+(0)(1)-1=0. $$ The partial derivatives are $$ F_x(x,y)=y,\qquad F_y(x,y)=3y^2+x. $$ At \((0,1)\), \(F_x(0,1)=1\) and \(F_y(0,1)=3(1)^2+0=3\ne0\). Hence $$ \left.\frac{dy}{dx}\right|_{(0,1)} =-\frac{1}{3}. $$
Differentiate the original equation along the curve to check: $$ 3y^2y'+y+xy'=0. $$ At the point, the left side is $$ 3(1)^2\left(-\frac13\right)+1+0\left(-\frac13\right) =-1+1+0=0. $$ So the computed slope satisfies the differentiated equation.
Why the Nonzero-Denominator Condition Matters
The formula requires \(F_y\ne0\) at the point. If \(F_y=0\), division by \(F_y\) is not valid, and the stated implicit function theorem does not guarantee a local differentiable representation \(y=y(x)\). This does not prove that no such representation exists; it means this condition and formula cannot settle the question at that point.
For example, consider \(F(x,y)=x^2+y^2\) at \((0,0)\). The point satisfies \(F(0,0)=0\), but \(F_y(0,0)=2(0)=0\). The equation \(x^2+y^2=0\) has only the point \((0,0)\) as a real solution, so it does not describe a curve that can be followed as a graph \(y(x)\) on an interval around \(x=0\). The quotient \(-F_x/F_y\) at that point would divide by zero.
Also, \(F_y\ne0\) is a condition for solving locally for \(y\) as a function of \(x\). If instead \(F_x\ne0\), the implicit function theorem may describe \(x\) locally as a function of \(y\). That is a different choice of dependent variable and is not the formula for \(dy/dx\) stated here.
A Common Mistake: Forgetting That \(y\) Depends on \(x\)
A frequent error is to differentiate an expression involving \(y\) as if \(y\) were constant. For the circle \(x^2+y^2=25\), an incorrect calculation might be $$ 2x+2y=0, $$ which would suggest \(y'=-1\). This has omitted the chain-rule factor \(y'\) in the derivative of \(y^2\).
Because \(y=y(x)\), the correct derivative is $$ 2x+2y\frac{dy}{dx}=0. $$ At \((3,4)\), solving gives $$ 6+8\frac{dy}{dx}=0,\qquad \frac{dy}{dx}=-\frac{6}{8}=-\frac34. $$ Substitution confirms the correction: \(6+8(-3/4)=0\). The missing factor is not optional; it records how quickly \(y\) changes as \(x\) changes.
What the Formula Tells You
The numerator \(F_x\) measures the change in \(F\) from changing \(x\) while holding \(y\) fixed. The denominator \(F_y\) measures the change in \(F\) from changing \(y\) while holding \(x\) fixed. Along the curve, the total change in \(F\) must be zero. The slope \(y'\) balances those two contributions, which is why it is the negative ratio \(-F_x/F_y\).
When using the formula, check three things in order:
- The point lies on the curve, so \(F(x,y)=0\).
- The partial derivative \(F_y\) is nonzero at the point.
- The partial derivatives are evaluated at the point on the curve when forming \(-F_x/F_y\).
These checks connect the algebraic calculation to the hypotheses that make the formula valid. The formula is not obtained by treating \(x\) and \(y\) as independent while differentiating the constraint; it comes from applying the chain rule to the local dependence \(y=y(x)\).
Check Your Understanding
For each question, identify the relevant partial derivatives and keep track of when the formula applies.
- In one sentence, explain why differentiating \(F(x,y(x))=0\) produces a term containing \(y'(x)\).
- For \(F(x,y)=x^2+3y^2-12\), find \(dy/dx\) at \((3,1)\), and verify the point is on the curve first.
- What does the condition \(F_y(a,b)\ne0\) guarantee in the implicit function theorem used here?
- For \(F(x,y)=x^2+y^2-1\), can the formula for \(dy/dx\) be used at \((1,0)\)? State why or why not.
- Describe the error in differentiating \(y^4\) along a curve as \(4y^3\) without an additional factor.