Composite and implicit functions

One parameter, one path. If $z=f(x,y)$ is differentiable and both $x=x(t)$, $y=y(t)$ trace out a path as $t$ varies, then $z$ becomes, indirectly, a function of $t$ alone, and its rate of change is a weighted sum of the two partial rates of $f$, weighted by how fast $x$ and $y$ themselves move:

where: $f_x=\partial f/\partial x$ and $f_y=\partial f/\partial y$ are the partial derivatives of $f$ held at the current point, and $\dot x=dx/dt$, $\dot y=dy/dt$ are the ordinary derivatives of the path coordinates. This is nothing but the total differential $df=f_x\,dx+f_y\,dy$ divided through by $dt$.

Several parameters, a dependency tree. If $z=f(x_1,\dots,x_n)$ and every $x_i$ is itself a function of variables $u_1,\dots,u_m$, the same idea sums over every route from $z$ down to a given $u_j$:

where: the sum runs over every intermediate variable $x_i$ that both depends on $u_j$ and feeds into $z$; skipping a branch of the tree is the most common way to lose a term. The picture below shows why: $z$ sits at the top, each $x_i$ is a middle node, each $u_j$ a bottom node, and $\partial z/\partial u_j$ adds up one product per path connecting top to bottom.

Implicit differentiation. Suppose $y$ is defined only implicitly by $F(x,y)=0$ — say $x^2+y^2-4=0$, which never mentions $y$ alone. Near any point where $F_y\neq0$ this equation still pins down $y$ as some function $y(x)$, even without an explicit formula, and the chain rule above (applied to $F(x,y(x))$, a composite of $x$ with itself and with $y(x)$) gives a shortcut for $dy/dx$ directly from the partial derivatives of $F$ — worked out fully in the derivation below.

Find the slope $dy/dx$ at the point $(1,2)$ on the curve defined implicitly by $F(x,y)=x^3+y^3-3xy-3=0$, then use that slope to write the tangent line at $(1,2)$.

Step 1 — verify the point lies on the curve. $F(1,2)=1^3+2^3-3(1)(2)-3=1+8-6-3=0$. Confirmed, so $(1,2)$ is a genuine point of the curve and the implicit-differentiation formula applies there.

Step 2 — compute the partial derivatives $F_x,F_y$.

Step 3 — evaluate at $(1,2)$.

Step 4 — apply the formula from the derivation.

So the curve has slope $\dfrac{1}{3}$ at $(1,2)$, found without ever solving the cubic relation $x^3+y^3-3xy-3=0$ explicitly for $y$. The slope at $(1,2)$ is $\mathbf{1/3}$.

Competitive stretch

With the slope in hand, the tangent line at $(1,2)$ is $y-2=\tfrac13(x-1)$, i.e. $3y-6=x-1$, i.e. $x-3y+5=0$. As a quick cross-check, compare with the single-parameter chain rule from the core idea: for $z=x^2+3xy$ with $x=\cos t,\ y=\sin t$, the same machinery gave $dz/dt|_{t=\pi/2}=(2x+3y)(-\sin t)+(3x)(\cos t)=-3$ at $t=\pi/2$ — a reminder that both the ordinary chain rule and implicit differentiation reduce to the same algebraic step of substituting partial derivatives into a linear combination. This pattern — differentiate an identity, then solve one algebraic equation for the rate you actually want — is the single most reusable trick in this part of the syllabus: it is exactly how the age-of-a-curve slope above was found, and it generalizes without change to surfaces $F(x,y,z)=0$, where the same steps give $\partial z/\partial x=-F_x/F_z$ and $\partial z/\partial y=-F_y/F_z$.

• Dropping a branch of the dependency tree. With three or more intermediate variables, every one of them contributes a term $f_{x_i}\cdot\partial x_i/\partial u_j$; skipping a branch silently throws away part of the answer.
• Writing $\partial z/\partial t$ instead of $dz/dt$. When $z=f(x,y)$ with $x,y$ functions of one variable $t$, the result is a total (ordinary) derivative, even though the calculation uses the partial derivatives of $f$.
• Forgetting $dy/dx$ on $y$-terms when differentiating $F(x,y)=0$ directly. Differentiating $y^3$ with respect to $x$ gives $3y^2\,dy/dx$, not $3y^2$ — every $y$-term needs the extra factor because $y=y(x)$.
• Using $dy/dx=-F_x/F_y$ without checking $F_y\neq0$. At points where $F_y=0$, the tangent is vertical and the formula is undefined — not a computational slip but a real geometric feature of the curve.

Ask the tutor: if a multi-step chain-rule computation with several intermediate variables gets tangled, or an implicit-differentiation slope is not simplifying the way you expect, ask the AI tutor to walk through the dependency tree with you step by step.