Nested Chain Rule and Implicit Differentiation Workshop

Choosing the Derivative Rule

A complicated expression does not always require a complicated method. The first task is to identify how the quantities depend on one another. If a function is defined by several nested substitutions, use the chain rule to follow each dependency. If variables are tied together by an equation rather than by an explicit formula, differentiate that equation while keeping track of which variables depend on which others.

In a mixed problem, the same calculation can involve both ideas. For example, an implicit equation may contain a nested expression such as \((x^2+y)^3\). Differentiating that term requires the chain rule, while differentiating the entire equation requires implicit differentiation. The dependable strategy is to name the intermediate expression, differentiate its layers, and only then evaluate at the point of interest.

Working principle: Before differentiating, state which quantities are independent and which are functions of others. Then apply the chain rule along every dependency path, and check the result in the differentiated defining equation.

Nested Compositions: Follow Each Input

Suppose \(z=G(u,v)\), where \(u=u(x,y)\) and \(v=v(x,y)\). Then \(z\) depends on \(x\) and \(y\) through both intermediate variables. The multivariable chain rule gives

\frac{\partial z}{\partial x} = G_u\frac{\partial u}{\partial x} + G_v\frac{\partial v}{\partial x}, \qquad \frac{\partial z}{\partial y} = G_u\frac{\partial u}{\partial y} + G_v\frac{\partial v}{\partial y}.

Here \(G_u\) and \(G_v\) are evaluated at the intermediate values \(u(x,y)\) and \(v(x,y)\). Each term represents one route by which a change in the independent variable can affect \(z\). For a deeper nesting, apply the same rule one layer at a time: differentiate the outer function with respect to each input, then multiply by the derivative of that input.

Worked Example: Partial Derivatives of a Two-Layer Composition

Let \(z=G(u,v)=u^2+3uv+v^2\), where \(u=x^2+y\) and \(v=xy\). Find \(z_x\) and \(z_y\) at \((x,y)=(1,2)\).

First evaluate the intermediate variables:

u=1^2+2=3, \qquad v=(1)(2)=2.

The outer partial derivatives are

G_u=2u+3v=2(3)+3(2)=12, \qquad G_v=3u+2v=3(3)+2(2)=13.

The inner partial derivatives are \(u_x=2x=2\), \(v_x=y=2\), \(u_y=1\), and \(v_y=x=1\) at the point. Therefore,

$$ z_x=G_u u_x+G_v v_x=(12)(2)+(13)(2)=50, $$

$$ z_y=G_u u_y+G_v v_y=(12)(1)+(13)(1)=25. $$

To check, substitute \(u\) and \(v\) into \(G\) and differentiate the resulting expression using the product rule. Its \(x\)-derivative is \(2u u_x+3(u_xv+uv_x)+2v v_x\), which at the point is \(12+30+8=50\). Its \(y\)-derivative is \(2u u_y+3(u_yv+uv_y)+2v v_y=6+15+4=25\). Both calculations agree.

Implicit Curves: Differentiate the Defining Equation

An implicit curve is specified by an equation \(F(x,y)=0\), rather than by an explicit formula for \(y\). Near a point where \(F\) is continuously differentiable and \(F_y\ne 0\), the implicit function theorem guarantees that the curve can locally be represented as a differentiable function \(y=y(x)\). Differentiating the identity \(F(x,y(x))=0\) by the chain rule gives

F_x+F_y\frac{dy}{dx}=0, \qquad \frac{dy}{dx}=-\frac{F_x}{F_y}.

The condition \(F_y\ne0\) matters: it permits division by \(F_y\) and, under the stated smoothness condition, ensures a local representation with \(y\) depending on \(x\). The partial derivative \(F_x\) treats \(y\) as fixed; the derivative \(dy/dx\) does not, because it describes how \(y\) changes along the curve.

Why the formula works: Along the curve, \(F(x,y(x))\) is constantly zero. By the chain rule, its derivative is \(F_x+F_y y'\). Since the derivative of the constant zero is zero, \(F_x+F_y y'=0\). Solving this equation for \(y'\), when \(F_y\ne0\), gives \(y'=-F_x/F_y\).

Worked Example: A Nested Implicit Curve

Consider the curve \((x^2+y)^3+xy=1\). Find \(dy/dx\) at \((1,0)\), and verify the result.

First check that the point is on the curve: \((1^2+0)^3+(1)(0)=1\). Write \(F(x,y)=(x^2+y)^3+xy-1\). Its partial derivatives, using the chain rule on the cube, are

F_x=3(x^2+y)^2(2x)+y, \qquad F_y=3(x^2+y)^2+x.

At \((1,0)\), these become \(F_x=3(1)^2(2)+0=6\) and \(F_y=3(1)^2+1=4\). Since \(F_y=4\ne0\), the local implicit derivative formula applies. Thus

\frac{dy}{dx}=-\frac{6}{4}=-\frac{3}{2}.

For a direct check, differentiate the defining equation while treating \(y\) as \(y(x)\):

3(x^2+y)^2\left(2x+\frac{dy}{dx}\right) +y+x\frac{dy}{dx}=0.

At \((1,0)\), substituting \(dy/dx=-3/2\) gives \(3(1)(2-3/2)+0+1(-3/2)=3/2-3/2=0\), as required.

Implicit Surfaces: Choose the Variable Being Solved For

For a surface given by \(F(x,y,z)=0\), one variable may be a function of the other two. If \(F\) is continuously differentiable near a point and \(F_z\ne0\) there, the implicit function theorem gives a local differentiable representation \(z=z(x,y)\). The chain rule applied to \(F(x,y,z(x,y))=0\) yields

F_x+F_z\frac{\partial z}{\partial x}=0, \qquad F_y+F_z\frac{\partial z}{\partial y}=0.

Consequently, \(z_x=-F_x/F_z\) and \(z_y=-F_y/F_z\) at points where \(F_z\ne0\). The partial \(z_x\) describes change in \(z\) as \(x\) varies while \(y\) is held fixed; \(z_y\) describes change as \(y\) varies while \(x\) is held fixed. If a different variable is to be represented as a function of the other two, the corresponding partial derivative must be nonzero instead.

Worked Example: Partial Derivatives on an Implicit Surface

The surface is defined by \((x^2+y+z)^2+xz=10\). Find \(z_x\) and \(z_y\) at \((1,1,1)\), where \(z\) is locally a function of \(x\) and \(y\).

Set \(F(x,y,z)=(x^2+y+z)^2+xz-10\). At the point, \(x^2+y+z=1+1+1=3\), and \(3^2+(1)(1)=10\), so the point lies on the surface. Differentiating the outer square and the remaining product gives

F_x=2(x^2+y+z)(2x)+z, \qquad F_y=2(x^2+y+z), \qquad F_z=2(x^2+y+z)+x.

At \((1,1,1)\), these values are \(F_x=2(3)(2)+1=13\), \(F_y=2(3)=6\), and \(F_z=2(3)+1=7\). Since \(F_z\ne0\), both implicit partial derivatives are available:

z_x=-\frac{F_x}{F_z}=-\frac{13}{7}, \qquad z_y=-\frac{F_y}{F_z}=-\frac{6}{7}.

To check \(z_x\), differentiate the surface equation with respect to \(x\), holding \(y\) fixed: \(2(x^2+y+z)(2x+z_x)+z+xz_x=0\). At the point and with \(z_x=-13/7\), the left side is \(6(2-13/7)+1-13/7=6/7-6/7=0\). For \(z_y\), differentiation with respect to \(y\), holding \(x\) fixed, gives \(2(x^2+y+z)(1+z_y)+xz_y=0\). Substitution yields \(6(1-6/7)-6/7=6/7-6/7=0\). Both results satisfy the differentiated equation.

A Reliable Mixed-Problem Workflow

When a problem combines nesting and implicit dependence, separate the decisions. First identify the independent variables and the dependent variable. Next determine whether the requested derivative concerns an explicit composition or a variable constrained by an equation. Then differentiate each nested expression carefully, keeping the dependence of any implicit variable visible. Only after the derivatives are formed should you substitute the point’s coordinates.

Identify the dependency structure.
For a composition, list the intermediate inputs. For an implicit equation, state which variable depends on the others.

Choose the correct derivative.
Use partial derivatives for independent inputs to a composition, or differentiate the constraint along the implicitly defined curve or surface.

Apply the chain rule at every nested layer.
Include the derivative of each inner expression, and include the derivative of the dependent variable when differentiating an implicit relation.

Check the conditions and the result.
For the usual implicit formulas, verify that the relevant partial derivative is nonzero. Substitute the answer into the differentiated constraint.

Common Mistake: Freezing a Dependent Variable

A frequent error is to differentiate an implicit equation as though every symbol other than the differentiation variable were constant. In the curve example, this would omit the \(dy/dx\) contribution from the nested term. A wrong calculation might say

\frac{d}{dx}(x^2+y)^3 \ \text{is}\ 3(x^2+y)^2(2x),

and then conclude from the equation that \(6=0\) at \((1,0)\), an immediate sign that something has gone wrong. The correct derivative is

\frac{d}{dx}(x^2+y(x))^3 = 3(x^2+y)^2\left(2x+\frac{dy}{dx}\right).

The inner expression changes through both \(x\) and \(y(x)\). Including both contributions gives the valid equation \(3(1)(2+y')+y'=0\), which solves to \(y'=-3/2\). A separate, related error is to compute the outer partial derivatives of a composition but forget to multiply by the derivatives of its intermediate inputs. Naming \(u\) and \(v\) before differentiating helps prevent both omissions.

Why the Checks Matter

In implicit differentiation, the answer is not merely a number produced by a formula: it must satisfy the relation obtained by differentiating the original constraint. That check catches missing chain-rule terms, sign errors, and incorrect choices of which variable is held fixed. In a nested composition, independently expanding the composition can provide a second route to the same derivative, as in the first example.

These methods also clarify when a calculation is justified. If the partial derivative used as a denominator is zero, the standard local formula cannot be applied at that point. This does not by itself prove that no local representation exists; it means that this particular theorem-based representation and division have not been justified there. Identify the condition before using the formula rather than treating it as an algebraic convenience.

Check Your Understanding

For each question, identify the variables being held fixed and state the rule or condition needed before calculating.

If \(w=H(a,b)\), \(a=x+y\), and \(b=x^2y\), what two terms make up \(w_x\)?
For an implicit curve \(F(x,y)=0\), what condition permits the formula \(dy/dx=-F_x/F_y\) near a point?
When differentiating \((x+y(x))^4\) with respect to \(x\), what is the derivative of the inner expression?
For \(F(x,y,z)=0\) with \(z=z(x,y)\), write the equation obtained by differentiating with respect to \(y\).
Why should an implicit derivative be substituted into the differentiated constraint as a check?