Varying One Coordinate at a Time
A function of several variables can change in different ways as its input moves. A partial derivative measures change in just one coordinate direction: the chosen coordinate varies while all the others stay fixed. This turns the calculation into an ordinary one-variable derivative, but it does not describe every possible way to approach the point. The distinction will matter when we discuss continuity.
Let \(U\) be an open subset of \(\mathbb{R}^m\), let \(f:U\to\mathbb{R}\), and fix \(a=(a_1,\ldots,a_m)\in U\). For \(j\in\{1,\ldots,m\}\), let \(e_j\) be the vector with \(1\) in coordinate \(j\) and \(0\) in every other coordinate. Since \(U\) is open, \(a+t e_j\) belongs to \(U\) for every \(t\) sufficiently close to \(0\). The values of \(f\) along this short coordinate line form a one-variable function.
For a function \(f(x,y)\), the partial derivative with respect to \(x\) varies \(x\) and holds \(y\) fixed; the partial derivative with respect to \(y\) does the reverse. The value of a partial derivative is a number associated with the point. When the partial derivative exists at every point of an open set, it can also be viewed as a function on that set.
Partial Derivatives as Derivatives of Slices
The definition is precisely an ordinary derivative applied to a coordinate slice. For fixed \(a\) and \(j\), define \(g_j(t)=f(a+t e_j)\) for \(t\) in a sufficiently small open interval about \(0\). Then the difference quotient defining \(\partial_j f(a)\) is the difference quotient defining \(g_j'(0)\).
Proof. Because \(U\) is open and contains \(a\), there is an \(r>0\) such that the open Euclidean ball \(B_r^{(2)}(a)\) is contained in \(U\). If \(|t|<r\), then \[ \|a+t e_j-a\|_2=|t|<r, \] so \(a+t e_j\in U\) and \(g_j(t)\) is defined. Also \(g_j(0)=f(a)\). For every nonzero \(t\) in this interval, \[ \frac{g_j(t)-g_j(0)}{t} = \frac{f(a+t e_j)-f(a)}{t}. \] The two difference quotients are therefore identical wherever they are defined near \(0\). Their limits as \(t\to0\) exist together and, when they exist, are equal. This proves the claim. \(\square\)
This characterization lets us use ordinary one-variable differentiation rules, provided their hypotheses hold for the slice. For example, if a slice is differentiable at \(0\), it is continuous there: its difference quotient stays bounded near \(0\), so multiplying that quotient by \(t\) shows that \(g_j(t)-g_j(0)\to0\). We will use this fact in proving the algebra rules.
Worked Example: A Polynomial in Two Variables
Let \(f(x,y)=x^2y+3xy^2-4y\). To find \(\partial f/\partial x\), hold \(y\) fixed and differentiate each term with respect to \(x\): \[ \frac{\partial f}{\partial x}(x,y)=2xy+3y^2. \] To find \(\partial f/\partial y\), hold \(x\) fixed: \[ \frac{\partial f}{\partial y}(x,y)=x^2+6xy-4. \] At \((1,-1)\), these values are \[ \frac{\partial f}{\partial x}(1,-1)=2(1)(-1)+3(-1)^2=-2+3=1 \] and \[ \frac{\partial f}{\partial y}(1,-1)=1^2+6(1)(-1)-4=1-6-4=-9. \] The different values reflect the fact that the two coordinate slices can change at different rates.
Worked Example: Exponential and Trigonometric Terms
Define \(F(x,y)=e^{xy}+\sin(x+y)\). Holding \(y\) fixed when differentiating with respect to \(x\), and holding \(x\) fixed when differentiating with respect to \(y\), gives \[ \frac{\partial F}{\partial x}(x,y)=y e^{xy}+\cos(x+y), \qquad \frac{\partial F}{\partial y}(x,y)=x e^{xy}+\cos(x+y). \] At \((0,\pi/2)\), the first partial derivative is \[ \frac{\partial F}{\partial x}(0,\pi/2) =(\pi/2)e^0+\cos(\pi/2)=\pi/2+0=\pi/2, \] whereas the second is \[ \frac{\partial F}{\partial y}(0,\pi/2) =0\cdot e^0+\cos(\pi/2)=0. \] In the first calculation \(y\) is fixed at \(\pi/2\); in the second calculation \(x\) is fixed at \(0\).
Algebra Rules and Their Hypotheses
The ordinary product and quotient rules also apply to partial derivatives. For the quotient rule, it is important to check where the quotient is defined. The following version assumes the denominator is nonzero throughout an open neighborhood of the point. That condition ensures the quotient itself is defined on a domain around the point, not merely at points on one coordinate line.
Proof. Write \(f_t=f(a+t e_j)\) and \(g_t=g(a+t e_j)\) for sufficiently small \(t\), and write \(f_0=f(a)\), \(g_0=g(a)\). Linearity follows by taking the limit of the identity \[ \frac{\alpha f_t+\beta g_t-(\alpha f_0+\beta g_0)}{t} = \alpha\frac{f_t-f_0}{t}+\beta\frac{g_t-g_0}{t}. \] The limits on the right exist by hypothesis, so the limit on the left exists and has the stated value.
For the product, when \(t\neq0\) use the identity \[ \frac{f_tg_t-f_0g_0}{t} = f_t\frac{g_t-g_0}{t}+g_0\frac{f_t-f_0}{t}. \] Existence of the partial derivatives of \(f\) and \(g\) implies that their coordinate slices are continuous at \(0\), by the slice characterization and the one-variable fact established above. Hence \(f_t\to f_0\). Taking limits in the identity gives \[ \partial_j(fg)(a) =f_0\,\partial_j g(a)+g_0\,\partial_j f(a), \] as required.
For the quotient, the neighborhood condition ensures \(g_t\neq0\) for all sufficiently small \(t\); also \(g_0\neq0\). For such \(t\neq0\), direct subtraction gives \[ \frac{f_t/g_t-f_0/g_0}{t} = \frac{g_0(f_t-f_0)-f_0(g_t-g_0)}{t\,g_tg_0}. \] The coordinate slices of \(f\) and \(g\) are continuous at \(0\), so \(g_t\to g_0\). The numerator after division by \(t\) tends to \(g_0\partial_j f(a)-f_0\partial_j g(a)\), while the denominator \(g_tg_0\) tends to \(g_0^2\neq0\). Thus the limit exists and equals the stated quotient. Since \(f/g\) is defined throughout \(V\), this limit is its partial derivative under the definition above. \(\square\)
Worked Example: A Quotient on a Neighborhood
Consider \(Q(x,y)=(x^2+y)/(1+xy)\). Its natural domain is the set where \(1+xy\neq0\), which is open because \((x,y)\mapsto1+xy\) is continuous. At \((1,0)\), the denominator equals \(1\). In fact, if \(|x-1|<1/4\) and \(|y|<1/4\), then \(|x|<5/4\), so \[ |xy|<\frac{5}{16} \quad\text{and consequently}\quad 1+xy>1-\frac{5}{16}=\frac{11}{16}>0. \] Thus the denominator is nonzero throughout this open neighborhood, and the quotient rule applies there.
Let \(N(x,y)=x^2+y\) and \(D(x,y)=1+xy\). The partial derivatives of the numerator and denominator are \(N_x=2x\), \(N_y=1\), \(D_x=y\), and \(D_y=x\). The quotient rule gives \[ Q_x(x,y)=\frac{2x(1+xy)-(x^2+y)y}{(1+xy)^2}, \qquad Q_y(x,y)=\frac{(1+xy)-(x^2+y)x}{(1+xy)^2}. \] At \((1,0)\), substitution yields \[ Q_x(1,0)=\frac{2(1)(1+0)-(1+0)0}{1^2}=2 \] and \[ Q_y(1,0)=\frac{1+0-(1+0)(1)}{1^2}=0. \] The neighborhood check is part of justifying the quotient rule here, not an optional detail.
What Partial Derivatives Do Not Tell Us
A partial derivative records change along one coordinate line. Even if every partial derivative exists at a point, those separate one-dimensional limits need not control values approached from other directions. In particular, existence of all partial derivatives at a point does not imply continuity there.
Worked Example: Partial Derivatives Without Continuity
Define \(q:\mathbb{R}^2\to\mathbb{R}\) by \[ q(x,y)= \begin{cases} \dfrac{xy}{x^2+y^2},&(x,y)\neq(0,0),\\ 0,&(x,y)=(0,0). \end{cases} \] At the origin, the \(x\)-coordinate slice is \(q(t,0)=0\) for every \(t\), including \(t=0\). Therefore \[ \frac{\partial q}{\partial x}(0,0) =\lim_{t\to0}\frac{q(t,0)-q(0,0)}{t} =\lim_{t\to0}0=0. \] Likewise, \(q(0,t)=0\) for every \(t\), so \[ \frac{\partial q}{\partial y}(0,0) =\lim_{t\to0}\frac{q(0,t)-q(0,0)}{t} =0. \] Nevertheless, for every nonzero \(t\), \[ q(t,t)=\frac{t^2}{t^2+t^2}=\frac12. \] Taking \(t=1/k\), the sequence \((1/k,1/k)\) converges to \((0,0)\), but the corresponding function values are all \(1/2\), not \(q(0,0)=0\). By the Sequential Criterion for Continuity in Euclidean Spaces, \(q\) is not continuous at the origin. Both partial derivatives exist there, yet the function is discontinuous.
The example also illustrates why checking the coordinate slices alone is insufficient for continuity: those slices both give value \(0\), while the diagonal gives value \(1/2\). Partial derivatives answer a narrower question. They measure local change along individual coordinate lines, not the full behavior of the function near the point. The upcoming study of directional derivatives will examine change along other straight-line directions; even those derivatives require careful interpretation and do not by themselves replace a continuity argument.
When computing a partial derivative, identify which variable is changing and treat the remaining coordinates as fixed. When using a quotient rule, verify that the quotient is defined on a neighborhood if the derivative is being defined for a function on an open domain. Finally, do not infer continuity merely from the existence of all coordinate partial derivatives; continuity requires control of the function as the full input approaches the point.
Check Your Understanding
Use the definition, the proved rules, and the examples to answer each question.
- How is the partial derivative with respect to coordinate \(j\) expressed as an ordinary derivative of a one-variable slice?
- In the product rule proof, why does \(f(a+t e_j)\) tend to \(f(a)\) as \(t\to0\)?
- What neighborhood condition on the denominator was used to justify the quotient rule as stated?
- What are the two partial derivatives of \(x^2y+3xy^2-4y\) at \((1,-1)\)?
- Why do the two partial derivatives of \(q\) at the origin fail to establish continuity there?