Partial Derivatives and Tangent Planes

The Essentials

In functions of multiple variables, a derivative doesn't make sense. A derivative is the instantaneous rate of change of a function, but in functions of two or more variables, there is not necessarily only one slope at any given point in the function. Instead of a tangent line to a curve, derivatives of multi-variable functions can be represented by a plane tangent to a surface. Given the two variable function \( f(x,y) \), the partial derivative of \( f \) with respect to \( x \) gives the slope of the tangent plane in the \( x \) direction. A partial derivative is found by setting the other variables to be constant.

When taking second order partial derivatives, taking it with respect to \( x \) first and then \( y \) is the same as taking it with respect to \( y \) first the \( x \), as long as the function is continuous:

\[ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)= \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) \]

These are standard convention for partial derivatives:

\[ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)= \frac{\partial^2f}{\partial x\partial y}= f_{yx} \]

The equation for the plane tangent to a function \( f(x,y) \) at the point \( P(x_0,y_0) \) is:

\[ z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) \]

The tangent plane can be used to approximate values of the function at the point \( P(x,y) \), given that the point \( P(x,y) \) is near where the plane is tangent to the function: \( P(x_0,y_0) \). This is accomplished by finding the equation for the tangent plane, then plugging in \( P(x,y) \).

The differential \( dz \) is called the total differential and is given by:

\[ dz=f_xdx+f_ydy \]

Example

Use a tangent plane of the function \( f(x,y)=x^2+2xy+y^2 \) at the point \( (3,2) \) to approximate the value of \( f \) at the point \( (3.1,2.1) \)

First we will take the partial derivatives of the function with respect to \( x \) and \( y \):

\[ f_x=2x+2yf_y=2x+2y \]

Next, we will find the plane tangent to the point (3, 2):

\[ z=f(3,2)+f_x(3,2)(x-3)+f_y(3,2)(y-2) \]
\[ z=25+10(x-3)+10(y-2) \]
\[ z=-25+10x+10y \]

Now we can plug in the point (3.1, 2.1) to approximate the value of the function:

\[ z=-25+31+21 \]
\[ z=27 \]

Which is close to the functions actual value at (3, 2) of \( z=25 \).

Practice

1) Find \( f_x \), \( f_y \), and \( f_z \) for the function: \( w=f(x,y,z)=\sin(x)y^2z+3\cos(z)-xe^y+3z^2-5yz+x+2 \)

2) Use the plane tangent to the function \( f(x,y) \) at the point \( (1,1) \) to approximate the value of the function at the point \( (1.1,0.9) \):

\[ f(x,y)=\frac{x}{y}+x^2-xy+2 \]

Solutions:

1)

\[ \begin{align*} f_x &= \frac{\partial}{\partial x}[\sin(x)y^2z] + \frac{\partial}{\partial x}[3\cos(z)] -\frac{\partial}{\partial x}[xe^y] + \frac{\partial}{\partial x}[3z^2]-\frac{\partial}{\partial x}[5yz] + \frac{\partial}{\partial x}[x] + \frac{\partial}{\partial x}[2] \\ &= \cos(x)y^2z + 0 - e^y + 0 - 0 + 1 + 0 \\ &= \cos(x)y^2z-e^y+1 \\ \\ f_y &= \frac{\partial}{\partial y}[\sin(x)y^2z] + \frac{\partial}{\partial y}[3\cos(z)] -\frac{\partial}{\partial y}[xe^y] + \frac{\partial}{\partial y}[3z^2]-\frac{\partial}{\partial y}[5yz] + \frac{\partial}{\partial y}[x] + \frac{\partial}{\partial y}[2] \\ &= 2\sin(x)yz + 0 - xe^y + 0 - 5z + 0 + 0 \\ &= 2\sin(x)yz - xe^y - 5z \\ \\ f_z &= \frac{\partial}{\partial z}[\sin(x)y^2z] + \frac{\partial}{\partial z}[3\cos(z)] -\frac{\partial}{\partial z}[xe^y] + \frac{\partial}{\partial z}[3z^2]-\frac{\partial}{\partial z}[5yz] + \frac{\partial}{\partial z}[x] + \frac{\partial}{\partial z}[2] \\ &= \sin(x)y^2 -3\sin(z) - 0 + 6z -5y + 0 + 0 \\ &= \sin(x)y^2 - 3\sin(z) + 6z - 5y \\ \end{align*} \]

2) 3.4

More Resources

Openstax: Partial Derivatives External Link Icon

Openstax: Tangent Planes and Linear Approximations External Link Icon

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