Partial derivatives tell us the slopes of two coordinate traces through a surface. To approximate the whole surface near a point, we need more: one plane must work for every nearby direction, not just along those two traces. That is what differentiability requires.

From a tangent line to a tangent plane

For a one-variable function, the tangent line at $x=a$ is the graph of

$$L(x)=f(a)+f'(a)(x-a).$$

Its value agrees with $f$ at $a$, and its slope agrees with $f'(a)$. But matching a value and a slope is not the whole story. The definition of the derivative says that the approximation error is small compared with the input change:

$$\begin{aligned} f'(a)&=\lim_{h\to0}\frac{f(a+h)-f(a)}{h},\\ 0&=\lim_{h\to0}\frac{f(a+h)-f(a)-f'(a)h}{h}. \end{aligned}$$

Equivalently, the absolute error divided by $|h|$ tends to zero.

A tangent line gives a first-order approximation

Now let $z=f(x,y)$ be a surface, and fix a point $(a,b)$ in an open region where $f$ is defined. The corresponding point on the surface is $(a,b,f(a,b))$. If the partial derivatives exist, they suggest a plane with slopes $f_x(a,b)$ and $f_y(a,b)$:

$$\boxed{L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b).}$$

Fixing $y=b$ gives the tangent line to the $y=b$ coordinate trace; fixing $x=a$ gives the tangent line to the $x=a$ trace. The candidate plane contains both lines. We still need to check whether it approximates the surface for changes in both variables together.

Differentiability means small relative error

Write $h=x-a$ and $k=y-b$ for the changes in the inputs. The actual output minus the candidate approximation is

$$\boxed{E(h,k)=f(a+h,b+k)-[f(a,b)+f_x(a,b)h+f_y(a,b)k].}$$

The function is differentiable at $(a,b)$ when its first partials exist there and

$$\boxed{\lim_{(h,k)\to(0,0)}\frac{|E(h,k)|}{\sqrt{h^2+k^2}}=0.}$$

The denominator is the distance moved in the input plane. So the condition says that the error becomes negligible compared with the size of the input change. Merely showing $E(h,k)\to0$ is weaker and is not enough. This is a two-variable limit: checking only the $h$- and $k$-axes cannot prove it.

When differentiability holds, $L$ is called the linearization, or linear approximation, at $(a,b)$. Its graph is the tangent plane:

$$\boxed{z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b).}$$

The coefficients are evaluated at the fixed base point, not at the nearby input. Near that point, we use $f(x,y)\approx L(x,y)$. The part $f_x(a,b)h+f_y(a,b)k$ is linear in the increments; the constant $f(a,b)$ places the plane at the correct height.

Differentiability also implies continuity. Indeed,

$$f(a+h,b+k)-f(a,b)=f_x(a,b)h+f_y(a,b)k+E(h,k).$$

The first two terms tend to zero. The last does too, because it is the input distance times a quantity tending to zero. Thus $f(a+h,b+k)\to f(a,b)$.

Example: a downward paraboloid

Find the tangent plane to

$$f(x,y)=-x^2-y^2$$

at $(1,1,-2)$. First check the height and compute the two slopes:

$$\begin{aligned} f(1,1)&=-2,\\ f_x(x,y)&=-2x,&f_x(1,1)&=-2,\\ f_y(x,y)&=-2y,&f_y(1,1)&=-2. \end{aligned}$$

The candidate linearization is therefore

$$\begin{aligned} L(x,y)&=-2-2(x-1)-2(y-1)\\ &=2-2x-2y. \end{aligned}$$

We can verify differentiability directly here, rather than just write down the plane. Expanding the true value gives

$$f(1+h,1+k)=-2-2h-2k-h^2-k^2.$$

The linear part is $-2-2h-2k$, so

$$E(h,k)=-h^2-k^2, \qquad \frac{|E(h,k)|}{\sqrt{h^2+k^2}} =\sqrt{h^2+k^2}\longrightarrow0.$$

The approximation works in every direction. The tangent plane is

$$\boxed{z=2-2x-2y}, \qquad\text{or equivalently}\qquad \boxed{2x+2y+z=2}.$$

A plane approximates the surface in both input directions

There is also an exact comparison between the surface and this plane:

$$f(x,y)-L(x,y)=-(x-1)^2-(y-1)^2.$$

The surface lies below the plane, touching it at $(1,1,-2)$. That sign is a useful check on the drawing and the algebra.

A normal vector to the tangent plane

Rearrange the general tangent-plane equation as

$$-f_x(a,b)(x-a)-f_y(a,b)(y-b)+z-f(a,b)=0.$$

Reading off the coefficients gives a normal vector

$$\boxed{\mathbf n=\langle-f_x(a,b),-f_y(a,b),1\rangle.}$$

Its negative is another valid normal. The coordinate-trace tangent directions are

$$\mathbf v_x=\langle1,0,f_x(a,b)\rangle, \qquad \mathbf v_y=\langle0,1,f_y(a,b)\rangle.$$

Their dot products with $\mathbf n$ are both zero. For the paraboloid example, $\mathbf n=\langle2,2,1\rangle$, which agrees with the coefficients of $2x+2y+z=2$.

When continuous partials are enough

Checking the error limit from scratch is not usually necessary. A standard sufficient condition is:

If $f_x$ and $f_y$ exist in an open neighborhood of $(a,b)$ and are continuous there, then $f$ is differentiable at $(a,b)$.

The neighborhood matters: knowing only two partial-derivative values at one point does not establish a good plane approximation. The condition is sufficient, not necessary; failure of continuity of a partial derivative does not by itself prove failure of differentiability.

For example,

$$f(x,y)=xy^3+xy+\sin(xy)$$

has

$$\begin{aligned} f_x(x,y)&=y^3+y+y\cos(xy),\\ f_y(x,y)&=3xy^2+x+x\cos(xy). \end{aligned}$$

Both partials are continuous on all of $\mathbb R^2$, so $f$ is differentiable everywhere. The same reasoning applies to the polynomial in the paraboloid example.

A cone has no tangent-plane approximation at its tip

Consider

$$f(x,y)=\sqrt{x^2+y^2}.$$

Its graph is the upper circular cone. It is continuous at the origin, but a sharp tip prevents a single first-order plane approximation there.

The cone's coordinate trace has a corner

The failure already appears in a coordinate trace. Along $y=0$, we have $f(x,0)=|x|$, so the candidate $x$-partial at the origin would be

$$\lim_{h\to0}\frac{f(h,0)-f(0,0)}h =\lim_{h\to0}\frac{|h|}{h}.$$

The quotient is $1$ for positive $h$ and $-1$ for negative $h$. The two-sided limit does not exist. Hence $f_x(0,0)$ does not exist, and $f$ cannot be differentiable there. The same argument works for $f_y(0,0)$.

Away from the origin, the partial derivatives are

$$f_x(x,y)=\frac{x}{\sqrt{x^2+y^2}}, \qquad f_y(x,y)=\frac{y}{\sqrt{x^2+y^2}}.$$

They are continuous wherever $(x,y)\ne(0,0)$, so the cone is differentiable away from its tip. This example separates continuity from differentiability; it is not an example with existing partial derivatives at the origin.