In one variable, the tangent line is the best straight-line approximation of a curve near a point. For a smooth surface, the best flat approximation is a tangent plane. You know how to write a plane from a point and a normal vector, and the two partial derivatives give you everything you need.
Key idea
The tangent plane to $z = f(x, y)$ at $(a, b, f(a, b))$ is
$$z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b).$$
Its right side is the linearization $L(x, y)$, and $f(x, y) \approx L(x, y)$ near $(a, b)$. A normal vector is $\langle f_x(a, b), f_y(a, b), -1 \rangle$.
Why it works
The plane must contain the tangent lines of both traces. Along the $x$-trace, a step of ${}1$ in $x$ raises $z$ by $f_x$, so that tangent line has direction $\langle 1, 0, f_x \rangle$; the $y$-trace gives $\langle 0, 1, f_y \rangle$. Their cross product is a normal:
$$\langle 1, 0, f_x \rangle \times \langle 0, 1, f_y \rangle = \langle -f_x, -f_y, 1 \rangle.$$
The plane through $(a, b, z_0)$ with this normal is $$-f_x(x - a) - f_y(y - b) + (z - z_0) = 0,$$ which rearranges to the formula above.
Worked examples
Worked example: A tangent plane
Find the tangent plane to $z = x^2 + 3xy - y^2$ at $(1, 2)$.
- $f(1, 2) = 1 + 6 - 4 = 3$.
- $f_x = 2x + 3y = 8$ and $f_y = 3x - 2y = -1$ at $(1, 2)$.
- $z = 3 + 8(x - 1) - (y - 2)$, which simplifies to $z = 8x - y - 3$.
Answer: $z = 8x - y - 3$
Worked example: An estimate
Use a linearization to estimate $\sqrt{(3.1)^2 + (3.9)^2}$.
- Let $f(x, y) = \sqrt{x^2 + y^2}$ near the friendly point $(3, 4)$, where $f = 5$.
- $f_x = \frac{x}{\sqrt{x^2 + y^2}} = \frac{3}{5}$ and $f_y = \frac{4}{5}$ there.
- $$L(3.1, 3.9) = 5 + \frac{3}{5}(0.1) + \frac{4}{5}(-0.1) = 5 + 0.06 - 0.08 = 4.98.$$
- A calculator gives ${}4.9820\ldots$: very close.
Answer: about ${}4.98$
Worked example: An exponential
Find the linearization of $f(x, y) = e^{x - y}$ at $(1, 1)$.
- $f(1, 1) = e^0 = 1$, $f_x = e^{x - y} = 1$, $f_y = -e^{x - y} = -1$ there.
- $$L = 1 + (x - 1) - (y - 1) = x - y + 1.$$
Answer: $L(x, y) = x - y + 1$
Common mistake
Leaving $x$ and $y$ inside the slopes. The coefficients in the tangent plane are the numbers $f_x(a, b)$ and $f_y(a, b)$. If your "plane" contains $x^2$ or $xy$, you forgot to evaluate them at the point. Check: a plane is linear in $x$, $y$, $z$, and it must pass through the point of tangency.