Deriving The First Implicit Differentiation Formula

From an Implicit Relation to a Derivative

An equation involving \(x\) and \(y\) can describe a relationship between the variables without specifying \(y\) as an explicit formula in terms of \(x\). Near a suitable point, the relationship may nevertheless define \(y\) as a differentiable function of \(x\). The key is to differentiate the equation while accounting for the fact that \(y\) changes as \(x\) changes.

Suppose a differentiable function \(F\) of two variables defines the relation \(F(x,y)=0\). If a differentiable branch of this relation is written as \(y=y(x)\), then along that branch the composite function \(F(x,y(x))\) is identically zero. Differentiating this composite gives the first implicit differentiation formula.

Central formula: Along any differentiable branch \(y=y(x)\) satisfying \(F(x,y(x))=0\), \[ F_x(x,y(x))+F_y(x,y(x))\frac{dy}{dx}=0. \] If \(F_y(x,y(x))\ne 0\), this equation can be solved for the derivative: \[ \frac{dy}{dx}=-\frac{F_x(x,y(x))}{F_y(x,y(x))}. \]

Partial Derivatives and the Chain Rule

Let \(F\) be a function of two variables, defined on an open subset of \(\mathbb{R}^2\). The partial derivative \(F_x(x,y)\) measures the rate of change of \(F\) with respect to \(x\), with \(y\) held fixed. The partial derivative \(F_y(x,y)\) measures the rate of change with respect to \(y\), with \(x\) held fixed.

F_x(x,y)=\frac{\partial F}{\partial x}(x,y), \qquad F_y(x,y)=\frac{\partial F}{\partial y}(x,y).

When \(y\) depends on \(x\), neither variable should be treated as fixed while following the curve. A small change in \(x\) changes the first input to \(F\) directly, and it can also change the second input through \(y(x)\). The chain rule adds these two contributions.

Chain rule for a two-variable composite: If \(F\) is differentiable and \(y=y(x)\) is differentiable, then \[ \frac{d}{dx}F(x,y(x)) = F_x(x,y(x))+F_y(x,y(x))\frac{dy}{dx}. \]

One way to see why the two terms appear is to use the first-order change in a differentiable function. When \(x\) changes by \(\Delta x\), the inputs change by \(\Delta x\) and \(\Delta y\), where \(\Delta y=y(x+\Delta x)-y(x)\). Differentiability of \(F\) gives the change in its output as the sum of its changes in each input, plus an error that is small compared with the size of the input change.

F(x+\Delta x,y+\Delta y)-F(x,y) = F_x(x,y)\Delta x+F_y(x,y)\Delta y +o\!\left(\sqrt{(\Delta x)^2+(\Delta y)^2}\right).

Because \(y\) is differentiable, \(\Delta y/\Delta x\) tends to \(dy/dx\) as \(\Delta x\) tends to zero. Divide the change in \(F\) by \(\Delta x\) and take the limit. The two displayed linear terms become \(F_x\) and \(F_y(dy/dx)\), respectively. The error term divided by \(\Delta x\) tends to zero: differentiability of \(y\) implies that \(\Delta y/\Delta x\) stays bounded near the point, so the size of the input change is bounded by a constant times \(|\Delta x|\). This yields the stated chain rule.

Deriving the First Implicit Differentiation Formula

Now suppose the branch \(y=y(x)\) satisfies the relation \(F(x,y(x))=0\) throughout an interval. The left side is a composite function of \(x\), and its value is constantly zero. Its derivative must therefore also be zero. Apply the chain rule to that composite:

\frac{d}{dx}F(x,y(x)) = F_x(x,y(x))+F_y(x,y(x))\frac{dy}{dx} = \frac{d}{dx}(0) = 0.

This proves the identity along the branch. It does not yet require division by \(F_y\). If \(F_y\ne0\) at the point under consideration, ordinary algebra allows us to isolate \(dy/dx\): subtract \(F_x\) from both sides, then divide by \(F_y\).

F_y\frac{dy}{dx}=-F_x \qquad\Longrightarrow\qquad \frac{dy}{dx}=-\frac{F_x}{F_y}.

Every quantity in this formula is evaluated at the point \((x,y(x))\) on the branch. The condition \(F_y\ne0\) is essential for this division. If \(F_y=0\), the chain-rule identity remains valid, but this algebraic step is not available; the identity alone does not determine the derivative by this quotient.

When Is There a Differentiable Branch?

The formula describes the derivative of a differentiable branch when one is present. The implicit function theorem gives a standard condition that ensures such a branch exists locally. Suppose \(F\) has continuous first partial derivatives near \((a,b)\), that \(F(a,b)=0\), and that \(F_y(a,b)\ne0\). Then, near \(x=a\), the equation \(F(x,y)=0\) determines a unique differentiable function \(y=g(x)\) with \(g(a)=b\). Its derivative is given by the implicit differentiation formula.

Local existence and derivative: If \(F\) has continuous first partial derivatives near \((a,b)\), \(F(a,b)=0\), and \(F_y(a,b)\ne0\), then the zero set of \(F\) near \((a,b)\) is the graph of a unique differentiable function \(y=g(x)\). At \(x=a\), \[ g'(a)=-\frac{F_x(a,b)}{F_y(a,b)}. \]

The nonzero partial derivative is a local condition at the point. On a branch, the same quotient formula applies at any nearby point where \(F_y\) remains nonzero. The theorem does not say that a branch cannot exist when \(F_y=0\); rather, this particular condition does not guarantee a graph of the stated form there.

Worked Examples

Worked Example: A Trigonometric Relation

Consider \(F(x,y)=\sin x+\cos y-1\) at \((\pi/6,\pi/3)\). First check that this point lies on the relation:

F\!\left(\frac{\pi}{6},\frac{\pi}{3}\right) = \sin\frac{\pi}{6}+\cos\frac{\pi}{3}-1 = \frac12+\frac12-1 = 0.

Compute the partial derivatives, differentiating with respect to one variable at a time:

F_x(x,y)=\cos x, \qquad F_y(x,y)=-\sin y.

At the specified point,

F_x\!\left(\frac{\pi}{6},\frac{\pi}{3}\right)=\frac{\sqrt3}{2}, \qquad F_y\!\left(\frac{\pi}{6},\frac{\pi}{3}\right)=-\frac{\sqrt3}{2}\ne0.

The derivative formula therefore gives

\left.\frac{dy}{dx}\right|_{(\pi/6,\pi/3)} = -\frac{\sqrt3/2}{-\sqrt3/2} = 1.

Check the result in the chain-rule identity. At this point its left side is

F_x+F_y\frac{dy}{dx} = \frac{\sqrt3}{2} + \left(-\frac{\sqrt3}{2}\right)(1) = 0.

The equation is satisfied, as it must be for a derivative along the relation.

Worked Example: A Relation with an Exponential

Let \(F(x,y)=x^2+e^y-1\), and consider the point \((0,0)\). The point satisfies the relation because

$$ F(0,0)=0^2+e^0-1=0+1-1=0. $$

Its partial derivatives are

F_x(x,y)=2x, \qquad F_y(x,y)=e^y.

At \((0,0)\), these values are

F_x(0,0)=2(0)=0, \qquad F_y(0,0)=e^0=1.

Since \(F_y(0,0)\ne0\), the formula applies:

\left.\frac{dy}{dx}\right|_{(0,0)} = -\frac{F_x(0,0)}{F_y(0,0)} = -\frac{0}{1} = 0.

Substitution into the differentiated relation checks the answer:

F_x(0,0)+F_y(0,0)\left.\frac{dy}{dx}\right|_{(0,0)} = 0+1(0) = 0.

Worked Example: A Logarithmic Relation

Take \(F(x,y)=\ln y+xy\), whose domain requires \(y>0\), and examine \((0,1)\). The point is in the domain and satisfies the relation:

F(0,1)=\ln 1+(0)(1)=0+0=0.

Differentiate with respect to each variable while holding the other fixed:

F_x(x,y)=y, \qquad F_y(x,y)=\frac{1}{y}+x.

At \((0,1)\),

F_x(0,1)=1, \qquad F_y(0,1)=\frac11+0=1\ne0.

Thus the derivative at the point is

\left.\frac{dy}{dx}\right|_{(0,1)} = -\frac{1}{1} = -1.

Checking the chain-rule equation gives

F_x(0,1)+F_y(0,1)\left.\frac{dy}{dx}\right|_{(0,1)} = 1+1(-1) = 0.

A Common Chain-Rule Mistake

A frequent error is to differentiate a term involving \(y\) as though \(y\) were constant, even though \(y\) depends on \(x\). For example, differentiating \(\cos(y(x))\) gives \(-\sin(y(x))\,dy/dx\), not just \(-\sin(y(x))\). The extra factor is the derivative of the inner function \(y(x)\).

Incorrect calculation: From \(\sin x+\cos y=1\), writing \[ \cos x-\sin y=0 \] omits the chain-rule factor. The correct differentiation is \[ \cos x-\sin y\frac{dy}{dx}=0. \] This is exactly \(F_x+F_y(dy/dx)=0\) for \(F(x,y)=\sin x+\cos y-1\).

Another point to keep straight is the distinction between the partial derivatives and the derivative along the branch. \(F_x\) and \(F_y\) describe changes in the two inputs separately; \(dy/dx\) describes how the inputs move together along the relation. The chain rule combines these quantities rather than treating them as interchangeable.

Why the Formula Is Useful

The formula lets us find the rate of change of a locally defined branch without first rearranging the original relation into an explicit expression for \(y\). Its derivation also explains exactly what the calculation depends on: the branch must be differentiable for the chain rule, and division by \(F_y\) requires \(F_y\ne0\). Keeping those conditions visible helps distinguish a valid derivative formula from an unjustified algebraic step.

Check Your Understanding

Answer each question using the definitions and conditions stated above.

In \(F_x(x,y)\), which variable is held fixed while taking the partial derivative?
Why does differentiating \(F(x,y(x))=0\) produce a term involving \(dy/dx\)?
State the chain-rule identity obtained from a differentiable branch of \(F(x,y)=0\).
What condition allows the identity to be rearranged as \(dy/dx=-F_x/F_y\)?
For \(F(x,y)=x^2+e^y-1\) at \((0,0)\), what are \(F_x\), \(F_y\), and the derivative of the local branch?
Why does \(F_y=0\) prevent the quotient formula from being used, without proving that no differentiable branch exists?

Deriving The First Implicit Differentiation Formula

What You'll Learn

From an Implicit Relation to a Derivative

Partial Derivatives and the Chain Rule

Deriving the First Implicit Differentiation Formula

When Is There a Differentiable Branch?