Partial derivatives and geometry

A function of one variable has a single slope at each point. A function of two variables, $z=f(x,y)$ , is a whole surface in space, and it can rise steeply in one direction while staying flat in another — so "the slope" needs to specify which direction. The partial derivative answers this by freezing every variable except one and differentiating the rest with the ordinary single-variable rules already known from school. It sounds like a small trick, but it is the entry point to all of multivariable calculus: tangent planes, directional behaviour, optimisation of surfaces, and eventually the chain rule and gradients all build on this one idea of "hold everything else fixed." Geometrically, each partial derivative is the slope of a curve carved out of the surface by a vertical slicing plane, so the algebra of freezing variables has an exact picture in terms of cutting a surface with a knife.

The core insight

The partial derivative $f_x(a,b)$ freezes $y=b$ and differentiates $f$ with respect to $x$ using ordinary rules, giving the slope of the trace curve where the vertical plane $y=b$ cuts the surface $z=f(x,y)$ .

The core idea

Definition by a limit. For $z=f(x,y)$ , the partial derivative with respect to $x$ at $(a,b)$ is defined exactly like the ordinary derivative, except only $x$ is allowed to move:

f_x(a,b) = \frac{\partial f}{\partial x}(a,b) = \lim_{h\to 0} \frac{f(a+h,b)-f(a,b)}{h}.

where: $h$ is a small increment in the $x$ -direction only, $y$ stays fixed at $b$ throughout the limit, and $f_x$ (also written $\partial f/\partial x$ ) is the resulting rate of change. Swapping the roles of $x$ and $y$ gives

f_y(a,b) = \frac{\partial f}{\partial y}(a,b) = \lim_{h\to 0} \frac{f(a,b+h)-f(a,b)}{h},

now with $x$ held fixed at $a$ . For a function of three variables $f(x,y,z)$ , the same freezing trick gives $f_x$ , $f_y$ , $f_z$ in turn, each obtained by treating the other two letters as plain constants.

Computing without the limit. In practice the limit is never evaluated directly. Because $y$ is a fixed number while computing $f_x$ , every ordinary differentiation rule — power rule, product rule, chain rule for single-variable expressions — applies with $y$ carried along as a constant coefficient. This is the whole computational content of the topic: partial differentiation of a two-variable function is nothing more than single-variable differentiation performed twice, once per variable, with the other variable "switched off."

Geometric interpretation. The graph $z=f(x,y)$ is a surface. Fixing $y=b$ slices it with the vertical plane $y=b$ , producing a curve on the surface called a trace, described by $z=f(x,b)$ — now a function of $x$ alone, since $b$ is a fixed number. The partial derivative $f_x(a,b)$ is exactly the ordinary slope of this trace curve at $x=a$ , measured in the plane $y=b$ . Symmetrically, fixing $x=a$ and slicing with the plane $x=a$ produces the trace $z=f(a,y)$ , whose slope at $y=b$ is $f_y(a,b)$ . So a single point on the surface carries two natural slopes — one along each coordinate direction — and together they describe how the surface tilts locally; the plane through both tangent lines is the tangent plane, developed once differentiability is introduced in the next topic.

Figure

A saddle-shaped surface sliced by two perpendicular vertical planes showing the two trace curves whose slopes are the partial derivatives

Derivation AComputing and reading both partial derivatives of a polynomial surface

The goal is to compute $f_x$ and $f_y$ for a genuinely mixed two-variable polynomial where both variables appear in the same term, evaluate both at a specific point, and read the numeric results as two different surface slopes at that point — the full cycle the syllabus expects, done carefully enough that the "freeze one variable" rule is applied term by term rather than remembered as a slogan.

Step 1. Write the function and identify what "freezing $y$ " means for each term. Let $f(x,y) = x^2y + 3xy^3 - y^2$ . To find $f_x$ , treat $y$ as a fixed number in every term: $x^2y$ becomes a constant $y$ times $x^2$ , whose $x$ -derivative is $2xy$ ; $3xy^3$ becomes a constant $3y^3$ times $x$ , whose $x$ -derivative is $3y^3$ ; and $-y^2$ is a pure constant once $y$ is frozen, so its $x$ -derivative is $0$ .

Step 2. Assemble $f_x$ . Adding the three pieces from Step 1 gives $f_x = 2xy + 3y^3 - 0 = 2xy+3y^3$ . Notice that $y$ never got differentiated — it only ever appears as a multiplier, exactly as the freezing rule demands.

Step 3. Repeat with $x$ frozen to find $f_y$ . Now $x^2y$ becomes a constant $x^2$ times $y$ , giving $x^2$ ; $3xy^3$ becomes a constant $3x$ times $y^3$ , giving $3x\cdot 3y^2=9xy^2$ ; and $-y^2$ differentiates normally to $-2y$ . Summing gives $f_y = x^2+9xy^2-2y$ .

Step 4. Evaluate both at the point $(1,-1)$ . Substituting $x=1,y=-1$ into $f_x=2xy+3y^3$ gives $f_x(1,-1)=2(1)(-1)+3(-1)^3=-2-3=-5$ . Substituting into $f_y=x^2+9xy^2-2y$ gives $f_y(1,-1)=(1)^2+9(1)(-1)^2-2(-1)=1+9+2=12$ .

Step 5. Read the two numbers as two slopes on the surface. At the point $(1,-1,f(1,-1))$ , moving in the $x$ -direction while $y$ stays fixed at $-1$ , the surface descends with slope $-5$ (the trace curve is falling). Moving instead in the $y$ -direction while $x$ stays fixed at $1$ , the surface rises steeply with slope $12$ . The two values are unrelated in size or sign precisely because they describe two different, perpendicular directions on the surface — neither one alone tells you what happens along, say, a diagonal direction. Verified: $f_x=2xy+3y^3\Rightarrow f_x(1,-1)=-5$ ; $f_y=x^2+9xy^2-2y\Rightarrow f_y(1,-1)=12$ .

Worked example

The surface $z=9-x^2-y^2$ (a downward paraboloid, like a dome) is sliced by the vertical plane $y=1$ . Find the slope of the resulting trace curve at $x=2$ , and give the equation of the tangent line to that trace.

Step 1. Identify the trace. Setting $y=1$ in $z=9-x^2-y^2$ gives $z=9-x^2-1=8-x^2$ , a downward parabola lying in the plane $y=1$ . Its slope at $x=2$ is exactly $f_x(2,1)$ for $f(x,y)=9-x^2-y^2$ .

Step 2. Compute $f_x$ by freezing $y$ . Since $y^2$ is constant once $y$ is frozen, $f_x=\dfrac{\partial}{\partial x}(9-x^2-y^2)=-2x$ .

Step 3. Evaluate at $(2,1)$ . $f_x(2,1)=-2(2)=-4$ . The trace $z=8-x^2$ has slope $-4$ at $x=2$ .

Step 4. Write the tangent line. At $x=2$ , $z=8-(2)^2=4$ , so the tangent to the trace at the point $(2,1,4)$ satisfies

z-4=-4(x-2) \quad\Longrightarrow\quad z=-4x+12,

valid along the fixed line $y=1$ in three-dimensional space. The trace has slope $-4$ at $x=2$ , with tangent line $z=-4x+12$ (along $y=1$ ).

Competitive stretch

The same paraboloid $z=9-x^2-y^2$ is sliced instead by $x=2$ ; find the slope of that trace at $y=1$ and compare it with the answer above. Freezing $x=2$ gives $z=9-4-y^2=5-y^2$ , so $f_y=-2y$ and $f_y(2,1)=-2$ . The two slopes at the same point, $f_x(2,1)=-4$ and $f_y(2,1)=-2$ , differ because the paraboloid is not radially symmetric in its rate of change along the two axes at this point — both are needed to describe the full local tilt of the dome.

Common traps

Letting the frozen variable "come alive." Differentiating $x^2y$ with respect to $x$ while accidentally treating $y$ as a hidden function of $x$ (product-rule reflex) gives the wrong answer $2xy+x^2y'$ ; $y$ is a plain constant during $f_x$ , full stop.
Swapping which slope is which. $f_x$ is the slope along the direction where $y$ is fixed (motion in $x$ ); it is easy to misread a trace picture and attach the wrong letter to the wrong direction.
Over-applying the product rule to mixed terms. In a term like $xy^3$ , the other variable is only a constant multiplier when differentiating — $\partial_x(xy^3)=y^3$ directly, with no product-rule machinery needed since $y^3$ is treated as a fixed coefficient.
Treating one partial derivative as the whole story. $f_x(a,b)$ alone only describes the slope along one direction; the full first-order tilt of the surface at a point needs both $f_x$ and $f_y$ together.

Ask the tutor: if a mixed-term partial derivative or a trace-curve tangent line isn't coming out right, ask the AI tutor to walk through a similar function step by step.

Why does freezing a variable make sense — isn't the surface still changing in every direction at once?

It is, but a partial derivative deliberately answers a narrower question: how fast does $z$ change if you move along just one coordinate axis, ignoring motion along the other? That restriction is what makes ordinary single-variable differentiation applicable at all — you cannot differentiate "in general" along an arbitrary direction without more machinery (that comes later, with directional derivatives and the gradient). Freezing one variable turns a two-variable problem into a one-variable one you already know how to solve exactly.

Do f sub x and f sub y together fully describe the tangent plane to the surface?

Almost — they are the two essential ingredients, but strictly you also need the surface to be differentiable at that point (a slightly stronger condition than just having both partial derivatives exist, covered in the very next topic). Once that holds, the tangent plane is built from exactly these two slopes: it is the unique plane through the surface point whose slope in the $x$ -direction matches $f_x$ and whose slope in the $y$ -direction matches $f_y$ .

Is there a shortcut for functions with more complicated terms, like ones involving sine or exponentials?

No new rule is needed — every single-variable technique (power rule, chain rule, product rule) still applies exactly as usual, with the other variable simply carried as a constant. For example, $\partial_x\!\left(\sin(xy)\right)=y\cos(xy)$ by the ordinary chain rule, since $y$ multiplies the inside of the sine and is treated as a constant factor throughout. The only discipline required is remembering, at every step, which letter is currently "switched off."

Partial derivatives and geometry

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads