Deriving the Tangent Plane to a Graph

From a Local Approximation to a Plane

A differentiable function of two variables changes, near a point, by an amount that is well described by a linear expression in the changes of its two inputs. For a graph $z=f(x,y)$, that linear expression produces a plane: the tangent plane at the point.

The plane’s location is determined by the value of the function at the base point. Its two slopes are determined by the partial derivatives there. Deriving the equation from differentiability explains why those are the right coefficients, rather than treating the formula as something to memorize.

Definition: Let $f$ be differentiable at an interior point $(a,b)$ of its domain. The tangent plane to the graph $z=f(x,y)$ at $(a,b,f(a,b))$ is the plane whose vertical difference from the graph is negligible compared with the distance from $(x,y)$ to $(a,b)$ as $(x,y)$ approaches $(a,b)$.

Here “negligible compared with distance” means that the vertical difference, divided by $\sqrt{(x-a)^2+(y-b)^2}$, tends to zero. This is the first-order contact supplied by differentiability. The point must be interior so that the function is defined in a neighborhood in which the two input coordinates can vary.

The Tangent-Plane Formula

Write the changes in the input coordinates as $h=x-a$ and $k=y-b$. Differentiability at $(a,b)$ means that the function can be expressed as its value plus a linear part and a remainder:

$$ f(a+h,b+k) = f(a,b)+f_x(a,b)h+f_y(a,b)k+r(h,k), $$

where the remainder satisfies

\lim_{(h,k)\to(0,0)} \frac{r(h,k)}{\sqrt{h^2+k^2}}=0.

The linear part describes a plane above the input point $(a+h,b+k)$. Replacing $h$ by $x-a$ and $k$ by $y-b$ gives the tangent-plane equation.

Theorem: If $f$ is differentiable at $(a,b)$, the tangent plane to $z=f(x,y)$ at $(a,b,f(a,b))$ is $$ z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b). $$

The point $(a,b,f(a,b))$ lies on this plane: substituting $x=a$ and $y=b$ makes both difference terms zero and gives $z=f(a,b)$. The two partial derivatives supply the plane’s slopes in the $x$- and $y$-directions.

Why the Partial Derivatives Are the Coefficients

The formula follows directly from the differentiability expansion. A short argument also shows why the coefficients of the plane must be the partial derivatives. Consider a plane through $(a,b,f(a,b))$ written with unspecified slopes $A$ and $B$:

$$ z=f(a,b)+A(x-a)+B(y-b). $$

Move only in the $x$-direction by setting $y=b$. The slope of the graph in that direction, obtained by a difference quotient, is $f_x(a,b)$. The slope of the proposed plane in that direction is $A$. For the plane to agree with the graph to first order, these slopes must agree. Moving only in the $y$-direction gives the corresponding condition on $B$.

Use the differentiability expansion.
Set $k=0$. Then $f(a+h,b)-f(a,b)=f_x(a,b)h+r(h,0)$, where $r(h,0)/|h|\to0$.

Take the $x$-direction slope.
Divide by $h$, for $h\ne0$, and let $h\to0$. The remainder quotient tends to zero, so the graph’s slope is $f_x(a,b)$. Thus $A=f_x(a,b)$.

Use the $y$-direction slope.
Set $h=0$, divide the expansion by $k\ne0$, and let $k\to0$. This gives the graph’s slope $f_y(a,b)$, so $B=f_y(a,b)$.

Substitute the coefficients.
The only possible first-order plane has $A=f_x(a,b)$ and $B=f_y(a,b)$, yielding the tangent-plane formula.

This argument identifies both coefficients, and the remainder condition shows the resulting plane really has first-order contact. In particular, the graph’s vertical difference from the plane is $r(h,k)$; its ratio to the input distance tends to zero by differentiability. Thus the plane is not just a surface with the right two coordinate slopes: its error becomes negligible relative to distance in every direction.

Geometric Meaning of the Coefficients

Fix $y=b$ and vary $x$. The graph’s cross-section has slope $f_x(a,b)$ at the base point, and the tangent plane has that same slope in the $x$-direction. Fixing $x=a$ instead shows that $f_y(a,b)$ is the plane’s slope in the $y$-direction.

The two coefficients together determine the plane’s orientation. Rearranging the equation into an implicit form gives

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

Consequently, a normal vector to the plane is $\langle f_x(a,b),f_y(a,b),-1\rangle$, or any nonzero scalar multiple. This describes the same plane in a different way: the partial derivatives are its coordinate-direction slopes, while a normal vector describes its orientation perpendicular to the plane.

Worked Examples

Worked Example: A Quadratic Graph

Find the tangent plane to $z=f(x,y)=x^2+xy+y^2$ at the point with input coordinates $(1,2)$.

First find the height of the graph at the point:

$$ f(1,2)=1^2+(1)(2)+2^2=1+2+4=7. $$

Differentiate with respect to $x$, holding $y$ fixed, and with respect to $y$, holding $x$ fixed:

f_x(x,y)=2x+y,\qquad f_y(x,y)=x+2y.

Evaluate both partial derivatives at $(1,2)$:

f_x(1,2)=2(1)+2=4,\qquad f_y(1,2)=1+2(2)=5.

Use the tangent-plane formula and then simplify:

\begin{aligned} z&=7+4(x-1)+5(y-2)\\ &=7+4x-4+5y-10\\ &=4x+5y-7. \end{aligned}

Check the point: at $x=1,y=2$, the plane gives $z=4+10-7=7=f(1,2)$. Its slopes in the two coordinate directions are $4$ and $5$, matching $f_x(1,2)$ and $f_y(1,2)$. As a further check of first-order contact, put $x=1+h$ and $y=2+k$. Direct expansion gives $f(1+h,2+k)=7+4h+5k+h^2+hk+k^2$; the plane gives $7+4h+5k$. Their difference is $h^2+hk+k^2$, whose ratio to $\sqrt{h^2+k^2}$ tends to zero.

Worked Example: An Exponential Graph

Find the tangent plane to $z=f(x,y)=e^{x-y}$ at $(0,0)$.

The height and partial derivatives are

f(0,0)=e^0=1,\qquad f_x(x,y)=e^{x-y},\qquad f_y(x,y)=-e^{x-y}.

At the base point, $f_x(0,0)=1$ and $f_y(0,0)=-1$. Therefore,

$$ z=1+1(x-0)-1(y-0)=1+x-y. $$

Substituting the base coordinates into the plane gives $z=1$, which equals $f(0,0)$. Its $x$-slope is $1$ and its $y$-slope is $-1$, matching the two partial derivatives there. The negative coefficient of $y$ reflects that, with $x$ held fixed, increasing $y$ decreases $e^{x-y}$.

Worked Example: The Upper Hemisphere

Find the tangent plane to $z=f(x,y)=\sqrt{9-x^2-y^2}$ at $(1,2)$. The point is in the interior of the domain because $1^2+2^2=5<9$.

Evaluate the function:

f(1,2)=\sqrt{9-1-4}=\sqrt{4}=2.

Differentiate and evaluate:

\begin{aligned} f_x(x,y)&=\frac{-2x}{2\sqrt{9-x^2-y^2}} =-\frac{x}{\sqrt{9-x^2-y^2}},\\ f_y(x,y)&=\frac{-2y}{2\sqrt{9-x^2-y^2}} =-\frac{y}{\sqrt{9-x^2-y^2}},\\ f_x(1,2)&=-\frac{1}{2},\qquad f_y(1,2)=-\frac{2}{2}=-1. \end{aligned}

Substitute these values into the formula and simplify:

\begin{aligned} z&=2-\frac12(x-1)-(y-2)\\ &=2-\frac{x}{2}+\frac12-y+2\\ &=\frac92-\frac{x}{2}-y. \end{aligned}

At $(1,2)$, the plane gives $9/2-1/2-2=2$, the correct height. The slopes are $-1/2$ in the $x$-direction and $-1$ in the $y$-direction, exactly the evaluated partial derivatives.

A Common Mistake: Using the Wrong Partial Derivative

For $f(x,y)=x^2+xy+y^2$, a frequent error is to differentiate $y^2$ with respect to $y$ as though its derivative were $y$, rather than $2y$. This leads to the incorrect calculation

f_y(x,y)\ \text{incorrectly taken as}\ x+y, \qquad f_y(1,2)\ \text{incorrectly taken as}\ 3.

That would give the wrong plane $z=7+4(x-1)+3(y-2)$. It passes through the correct point, but passing through the point alone does not make it tangent. To check the slope in the $y$-direction, hold $x=1$ fixed and calculate the difference quotient:

\frac{f(1,2+h)-f(1,2)}{h} = \frac{(1+2+h+(2+h)^2)-7}{h} = \frac{5h+h^2}{h} = 5+h.

As $h\to0$, this quotient tends to $5$, not $3$. The correct derivative is $f_y(x,y)=x+2y$, so $f_y(1,2)=5$, as used in the correct tangent plane. Always compute each partial derivative with respect to its own variable, then evaluate it at the base point.

Why the Tangent Plane Matters

The tangent plane provides a geometric picture of differentiability: sufficiently close to the base point, the graph is represented to first order by a plane. Its height at a nearby input is given by the plane equation, and the two partial derivatives explain how that plane tilts. The remainder condition is essential to this interpretation. Having partial derivatives at a point gives candidate slopes, but by itself does not guarantee that a function is differentiable there or that a tangent-plane approximation is valid.

When using the formula, keep the three pieces distinct: the height $f(a,b)$, the $x$-slope $f_x(a,b)$, and the $y$-slope $f_y(a,b)$. The resulting plane must pass through the graph point and have both coordinate-direction slopes. These checks catch many substitution and sign errors.

Check Your Understanding

Use the tangent-plane formula and the geometric interpretation to answer these questions.

What point on the graph must the tangent plane at $(a,b)$ contain?
In the tangent-plane equation, what geometric role does $f_x(a,b)$ play?
For $f(x,y)=x^2+y^2$, find the tangent plane at $(1,1)$, and check that it contains the graph point.
Why does differentiability ensure that the graph’s difference from its tangent plane is negligible relative to input distance?
Why are partial derivatives alone not enough to conclude that a tangent-plane approximation is valid?