The Chain Rule & Implicit Differentiation

Draw the dependency tree ($z$ on top, $x$ and $y$ below, $t$ at the bottom): multiply derivatives ALONG each path, add ACROSS paths. The rule extends to any number of variables and any depth of nesting — partials with respect to several inputs just select which bottom variable you follow. Intuition: a small change $dt$ perturbs $x$ by $x'dt$ and $y$ by $y'dt$; the tangent-plane approximation converts each into a $z$-change, and to first order the two effects simply add.

This retires the ad-hoc implicit differentiation of first-year calculus: compute two partials, form the ratio, attach a minus sign. In three variables, a surface $F(x,y,z) = 0$ has $\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z}$ by the same argument, and $\nabla F$ is the surface normal — tangent planes to implicit surfaces come free.

Why: hold $t$ fixed and let $s$ vary. Then $x$ and $y$ move along a curve, and the one-parameter chain rule from the first section applies with $s$ playing the role of time. The matrix form shows what the rule really says: a derivative is a linear map, and the derivative of a composition is the product of the matrices — the same statement as $(f\circ g)' = f'(g)\,g'$ in one variable, one size up.

So $\tfrac{\partial z}{\partial s} = 12\cdot1 + 9\cdot2 = 30$: $12$ arrives through $x$ and $18$ through $y$. Check by substituting first: $z = (s + t)^2\,st$, and $\tfrac{\partial z}{\partial s} = 2(s + t)\,st + (s + t)^2\,t = 2\cdot3\cdot2 + 9\cdot2 = 30$. In the same way $\tfrac{\partial z}{\partial t} = 12\cdot1 + 9\cdot1 = 21$, and directly $2(s + t)\,st + (s + t)^2\,s = 12 + 9 = 21$.

Substituting, $\dfrac{dw}{dt} = -\sin^2t + \cos^2t + 2t = \cos 2t + 2t$, which is $1$ at $t = 0$. Check: along the curve $w(t) = \sin t\cos t + t^2 = \tfrac12\sin 2t + t^2$, whose derivative is $\cos 2t + 2t$.

Take $f = xy$, so $f_x = y$ and $f_y = x$: then $f_r = y\cos\theta + x\sin\theta = 2r\sin\theta\cos\theta = r\sin 2\theta$. Directly, $f = r^2\sin\theta\cos\theta = \tfrac12 r^2\sin 2\theta$, and $\tfrac{\partial}{\partial r}$ gives $r\sin 2\theta$. Notice that $f_r = \nabla f\cdot(\cos\theta, \sin\theta)$ is the directional derivative straight away from the origin: the chain rule has rediscovered the gradient.

Why: hold $y$ fixed and differentiate $F(x, y, z(x, y)) = 0$ with respect to $x$. There are two routes from $F$ to $x$ — directly, and through $z$ — so the chain rule gives $F_x + F_z\,\tfrac{\partial z}{\partial x} = 0$, and solving gives the formula. The condition $F_z \ne 0$ is exactly what allows the solving: where $F_z = 0$ the surface stands vertical and $z$ is not locally a function of $x$ and $y$. The implicit function theorem makes this precise.

The gradient $(4, 5)$ is normal to the curve, so the tangent line is $4(x - 1) + 5(y - 2) = 0$, that is, $4x + 5y = 14$. Solving the curve for $y$ would take the quadratic formula; the implicit route took two partial derivatives.

So near $(1, 1, 2)$, $z \approx 2 - \tfrac{2}{13}(\Delta x + \Delta y)$. At $(1.1, 1)$ this predicts $z \approx 2 - \tfrac{0.2}{13} \approx 1.98462$, and the true root of $1.1z + z^3 = 10$ agrees to five decimal places. The value of $z$ was never solved for in general — a cubic — yet its rate of change came out in one line.

Multiply: $\left(-\tfrac{RT}{V^2}\right)\tfrac{R}{P}\tfrac{V}{R} = -\tfrac{RT}{PV} = -1$, using $PV = RT$. The product is $-1$, not the $+1$ that "cancelling" would suggest, because each factor holds a different variable fixed. It is $-1$ for every relation $F(P, V, T) = 0$: each factor is $-\tfrac{F_a}{F_b}$, the $F$'s cancel in pairs, and $(-1)^3 = -1$ remains.

As $\Delta t \to 0$ the difference quotients converge to $x'(t), y'(t)$, and $\tfrac{\varepsilon}{\Delta t} \to 0$ because $\varepsilon$ is sub-linear in the displacement. What survives is exactly the chain rule. $\blacksquare$ SIGNIFICANCE: every related-rates problem, every implicit derivative, every Jacobian change of variables, and every gradient-descent step is this limit argument reused.

$\blacksquare$ SIGNIFICANCE: in economics this says total output equals the sum of inputs paid their marginal products (constant returns to scale); in thermodynamics it generates the Gibbs–Duhem relation — one chain-rule differentiation, two sciences.

Step 1 — $V = \pi r^2h$ depends on two moving inputs, so $\dfrac{dV}{dt} = V_r\dfrac{dr}{dt} + V_h\dfrac{dh}{dt}$.

Step 2 — partials: $V_r = 2\pi rh = 30\pi$ and $V_h = \pi r^2 = 9\pi$ at the given moment.

Step 3 — plug in the rates: $\dfrac{dV}{dt} = 30\pi(2) + 9\pi(-1) = 51\pi$ cm$^3$/s.

Step 4 — interpret: the radius term dominates because volume is quadratic in $r$ but only linear in $h$ — the chain rule made that visible as the sizes of the two partials.

Step 1 — write the surface as $F(x,y,z) = x^2 + y^2 + z^2 - 14 = 0$; the normal to a level surface is the gradient $\nabla F = (2x, 2y, 2z)$.

Step 2 — evaluate: $\nabla F(1,2,3) = (2, 4, 6)$, or just $(1, 2, 3)$ after scaling.

Step 3 — the plane through the point with that normal: $1(x - 1) + 2(y - 2) + 3(z - 3) = 0$, i.e. $x + 2y + 3z = 14$.

Step 4 — sanity: for a sphere centered at the origin the normal at a point IS the point's position vector, and $14$ on the right equals $|(1,2,3)|^2$ — both facts confirm the plane without redoing anything.