# How to Find Tangent Planes of Multivariable Functions

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Updated: May 8, 2016

Finding tangent planes is a logical extension of finding equations of tangent lines on single-variable functions. Instead of computing ordinary derivatives, however, we compute gradients instead. Therefore, the goal is to arrive at a tangent plane of the form ${\displaystyle z-z_{0}={\frac {\partial z}{\partial x}}(x-x_{0})+{\frac {\partial z}{\partial y}}(y-y_{0}).}$

## Steps

1. 1
Begin with the function. Let's start with a function defined below. For this example, we want to find the tangent plane at the point ${\displaystyle (x_{0},y_{0},z_{0})=(1,2,4).}$
• ${\displaystyle z(x,y)=2x^{2}y^{3}-3x^{4}y^{2}}$
2. 2
Calculate the gradient of the function. Calculating the gradient is a higher-dimensional analog of calculating the derivative. The concept here is that we take partial derivatives to analyze the function in terms of how its ${\displaystyle x}$ and ${\displaystyle y}$ components individually change. The ${\displaystyle \nabla }$ symbol acting on a scalar function such as ${\displaystyle z}$ signals the presence of the gradient.
• In two-dimensional Cartesian coordinates, the gradient can be written as ${\displaystyle \nabla z=\left({\frac {\partial z}{\partial x}},{\frac {\partial z}{\partial y}}\right).}$ Remember that when taking partial derivatives with respect to some variable, the other variables in the function hold constant.
• ${\displaystyle {\frac {\partial z}{\partial x}}=4xy^{3}-12x^{3}y^{2}}$
• ${\displaystyle {\frac {\partial z}{\partial y}}=6x^{2}y^{2}-6x^{4}y}$
3. 3
Evaluate the gradient at the point. This is the same as evaluating a derivative at a particular point. We simply substitute the ${\displaystyle x_{0},y_{0}}$ values into the gradient and evaluate.
• ${\displaystyle {\frac {\partial z}{\partial x}}{\bigg |}_{(x_{0},y_{0})=(1,2)}=4(1)(2)^{3}-12(1)^{3}(2)^{2}=-16}$
• ${\displaystyle {\frac {\partial z}{\partial y}}{\bigg |}_{(x_{0},y_{0})=(1,2)}=6(1)^{2}(2)^{2}-6(1)^{4}(2)=12}$
4. 4
Substitute the values into the tangent plane equation. We now have all the information necessary to write the tangent plane of interest: the initial point, and how the function changes at that point. The general equation is restated below for your convenience.
• {\displaystyle {\begin{aligned}z-z_{0}&={\frac {\partial z}{\partial x}}(x-x_{0})+{\frac {\partial z}{\partial y}}(y-y_{0})\\z-4&=-16(x-1)+12(y-2)\end{aligned}}}

## Community Q&A

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## Tips

• Just as in the single-variable case, for a gradient to be evaluated at a particular point, the original function must be differentiable there as well.
• For example, consider the function ${\displaystyle z(x,y)={\frac {y-x}{x^{2}+y^{2}}}.}$
• The function ${\displaystyle z}$ is not continuous at the origin, so it is not differentiable there, and therefore the gradient cannot be evaluated at that point. Points like these are called singularities.
Thanks!
• Do not be confused by the different tasks that the variables perform in the example above. In the last step, they are written as part of the equation of the plane. In contrast, when calculating the gradient of the function ${\displaystyle z,}$ they are used to denote how that function changes. This possible confusion can easily be avoided if the gradient is calculated first and promptly evaluated at the point.
Thanks!

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