A partial derivative is an ordinary derivative with a twist. You freeze every variable but one, then take the derivative with respect to that one variable alone. Stack all of a function's partials into one vector and you get the gradient, \nabla f. It points in the direction f increases fastest. The chain rule, in its multivariable form, is the rule for taking the derivative of a chain of several such functions. It tracks how a change in the innermost variable ripples through every layer, all the way to the final output. Chapter 02 reframes all three ideas as one linear-map statement instead of three separate rules. Getting them solid now is what makes that payoff land.
Picture standing on a hillside. The gradient is the compass direction that takes you uphill fastest, right now, from exactly where you're standing. Walk perpendicular to it instead, and the ground stays level. That's the level-curve fact, wearing a hiking metaphor. The chain rule, here, is just tracking every route a wiggle can take to reach the final number. Say you wiggle east-west. Two things matter: how fast you climb per east-west step, and how fast east-west itself is changing over time. Multiply those two together, and that's the chain rule.
Partial derivatives work by treating every variable except one as a fixed constant. You then take the ordinary derivative of what's left: f_x freezes y and takes the derivative with respect to x; f_y does the same, holding x fixed instead. The gradient \nabla f = \langle f_x, f_y\rangle packages both partials into one vector. Its defining property is that it is perpendicular to level curves and points toward steepest increase. That property is really a statement about the directional derivative D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u}: it is maximized exactly when \mathbf{u} points along \nabla f itself, since the dot product is largest when two vectors align. The multivariable chain rule adds up each intermediate variable's contribution. If z=f(x,y) with x=x(t) and y=y(t), then \frac{dz}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt} — one term per path the change can travel. Chapter 02 reframes this whole picture as Df, a linear map. In that picture \nabla f is nothing more than that map's matrix.
Formula
The Gradient Vector
also known as: \nabla f, grad f, del f
Recalled from Calculus III — Multivariable Calculus · The gradient vector — first taught there.
For a differentiable function f(x,y), the gradient of f is the vector-valued function \nabla f(x,y) = \langle f_x(x,y),\, f_y(x,y)\rangle.
The directional derivative in the direction of a unit vector \mathbf{u} is then exactly the dot product of \mathbf{u} with the gradient: D_{\mathbf{u}}f(x,y) = \nabla f(x,y)\cdot\mathbf{u}.
Write \mathbf{u}=\langle a,b\rangle. The directional-derivative formula already gives D_{\mathbf{u}}f = f_x a + f_y b, and \langle f_x,f_y\rangle\cdot\langle a,b\rangle = f_x a + f_y b by the definition of the dot product — the two expressions are the same number, just written two different ways.
Formula
The Multivariable Chain Rule
also known as: chain rule for z = f(x,y) with x, y functions of t
Recalled from Calculus III — Multivariable Calculus · The multivariable chain rule — first taught there.
If z=f(x,y) is differentiable and x=x(t), y=y(t) are both differentiable functions of t, then z is a differentiable function of t and \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}. More generally, if x=x(s,t) and y=y(s,t) then \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}, \qquad \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}.
The gradient always points uphill, perpendicular to the contours
Drag the direction dial around the marked point -- the directional derivative peaks exactly when the dial lines up with the drawn gradient arrow.
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Slice the surface to see one partial derivative
Toggle the y=const trace plane -- its slope where it crosses the surface at the marked point is exactly f_x there.
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Worked example
Gradient and one partial
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1
For f(x,y) = x^2 y - y^3, find \nabla f and evaluate it at (1,1).
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2
f_x = 2xy, f_y = x^2 - 3y^2.
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3
\nabla f = \langle 2xy,\ x^2-3y^2\rangle.
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4
\boxed{\nabla f(1,1) = \langle 2,\ -2\rangle} — steepest ascent at (1,1) points toward increasing x and decreasing y.
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Where this goes next
Chapter 2 of this course
Every partial-derivative and gradient computation reviewed here survives unchanged into chapter 02. What changes is the reason it works. "The gradient is perpendicular to level curves" stops being a fact you check case by case. Instead, it falls out in one line from a single idea: the derivative is the best straight-line approximation to a function.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For f(x,y) = \sin(xy), find f_x(x,y).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Chain rule with y frozen as a constant: derivative of \sin(u) is \cos(u)\cdot u', where u=xy and u_x = y.
Correct answer
Also accepted:
Why
Chain rule: d/dx sin(xy) = cos(xy) * d/dx(xy).
From Calculus III — Multivariable Calculus
For f(x,y) = \dfrac{x}{x+y}, find f_x(1,1).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Quotient rule with y fixed: f_x = \dfrac{(x+y)(1) - x(1)}{(x+y)^2} = \dfrac{y}{(x+y)^2}, so f_x(1,1) = \dfrac{1}{4} = 0.25.
Correct answer
Also accepted:
Why
Use the quotient rule treating y as a constant.
f_x = y / (x+y)^2.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
Find \nabla f(1,2) for f(x,y)=\ln(x^2+y^2).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Enter as components — <2, -1, 3>, (2, -1, 3), or 2i - j + 3k all count. Any algebraically equivalent component counts too.
Why
At (1,2), x^2+y^2=5, so f_x=2/5=0.4 and f_y=4/5=0.8.
Correct answer
Also accepted:
Why
f_x=\dfrac{2x}{x^2+y^2}, f_y=\dfrac{2y}{x^2+y^2}.
From Calculus III — Multivariable Calculus
At a point where \nabla f=\langle0,0\rangle, what is D_{\mathbf{u}}f for every unit vector \mathbf{u}?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
The dot product of the zero vector with any vector is 0, so D_{\mathbf u}f=0 regardless of direction.
Correct answer
Also accepted:
Why
Use D_{\mathbf u}f=\nabla f\cdot\mathbf u directly.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For z=e^{xy}, x=t^2, y=t, find dz/dt at t=1.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
z_x=ye^{xy}, z_y=xe^{xy}, dx/dt=2t, dy/dt=1. dz/dt = ye^{xy}(2t)+xe^{xy}(1). At t=1: x=1,y=1,e^{xy}=e. dz/dt=1e2+1e1=3e approx 8.15.
Correct answer
Also accepted:
Why
At t=1, x=1 and y=1, so xy=1 and e^{xy}=e.
From Calculus III — Multivariable Calculus
For z=f(x,y) with x=x(t), y=y(t), the chain rule dz/dt = z_x*(dx/dt) + z_y*(dy/dt) sums:
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
z feels t through BOTH x changing and y changing, so both contributions are added.