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2.2: Partial Derivatives

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Now that we have an idea of what functions of several variables are, and what a limit of such a function is, we can start to develop an idea of a derivative of a function of two or more variables. We will start with the notion of a partial derivative.

Definition 2.3

Let f(x,y) be a real-valued function with domain D in R2, and let (a,b) be a point in D. Then the partial derivative of f at (a,b) with respect to x, denoted by fx(a,b), is defined as

fx(a,b)=limh0f(a+h,b)f(a,b)h

and the partial derivative of f at (a,b) with respect to y, denoted by fy(a,b), is defined as

fx(a,b)=limh0f(a+h,b)f(a,b)h

Note: The symbol is pronounced “del”.

Recall that the derivative of a function f(x) can be interpreted as the rate of change of that function in the (positive) x direction. From the definitions above, we can see that the partial derivative of a function f(x,y) with respect to x or y is the rate of change of f(x,y) in the (positive) x or y direction, respectively. What this means is that the partial derivative of a function f(x,y) with respect to x can be calculated by treating the y variable as a constant, and then simply differentiating f(x,y) as if it were a function of x alone, using the usual rules from single-variable calculus. Likewise, the partial derivative of f(x,y) with respect to y is obtained by treating the x variable as a constant and then differentiating f(x,y) as if it were a function of y alone.

Example 2.10

Find fx(x,y) and fy(x,y) for the function f(x,y)=x2y+y3.

Solution

Treating y as a constant and differentiating f(x,y) with respect to x gives

fx(x,y)=2xy

and treating x as a constant and differentiating f(x,y) with respect to y gives

fy(x,y)=x2+3y2

We will often simply write fx and fy instead of fx(x,y) and fy(x,y).

Example 2.11

Find fx and fy for the function f(x,y)=sin(xy2)x2+1.

Solution

Treating y as a constant and differentiating f(x,y) with respect to x gives

fx=(x2+1)(y2cos(xy2))(2x)sin(xy2)(x2+1)2

and treating x as a constant and differentiating f(x,y) with respect to y gives

fy=2xycos(xy2)x2+1

Since both fx and fy are themselves functions of x and y, we can take their partial derivatives with respect to x and y. This yields the higher-order partial derivatives:

2fx2=x(fx)2fy2=y(fy)

2fyx=y(fx)2fxy=x(fy)

3fx3=x(2fx2)3fy3=y(2fy2)

3fyx2=y(2fx2)3fxy2=x(2fy2)

3fy2x=y(2fyx)3fx2y=x(2fxy)

3fxyx=x(2fyx)3fyxy=y(2fxy)

Example 2.12

Find the partial derivatives fx, fy, 2fx2, 2fy2, 2fyx and 2fxy for the function f(x,y)=ex2y+xy3.

Solution

Proceeding as before, we have

fx=2xyex2y+y32fx2=x(2xyex2y+y3)=2yex2y+4x2y2ex2y2fyx=y(2xyex2y+y3)=2xex2y+2x3yex2y+3y2fy=x2ex2y+3xy22fy2=y(x2ex2y+3xy2)=x4ex2y+6xy2fxy=x(x2ex2y+3xy2)=2xex2y+2x3yex2y+3y2

Higher-order partial derivatives that are taken with respect to different variables, such as 2fyx and 2fxy, are called mixed partial derivatives. Notice in the above example that 2fyx=2fxy. It turns that this will usually be the case. Specifically, whenever both 2fyx and 2fxy are continuous at a point (a,b), then they are equal at that point. All the functions we will deal with will have continuous partial derivatives of all orders, so you can assume in the remainder of the text that

2fyx=2fxy for all (x,y) in the domain of f

In other words, it doesn’t matter in which order you take partial derivatives. This applies even to mixed partial derivatives of order 3 or higher.

The notation for partial derivatives varies. All of the following are equivalent:

fx:fx(x,y),f1(x,y),Dx(x,y),D1(x,y)

fy:fy(x,y),f2(x,y),Dy(x,y),D2(x,y)

2fx2:fxx(x,y),f11(x,y),Dxx(x,y),D11(x,y)

2fy2:fyy(x,y),f22(x,y),Dyy(x,y),D22(x,y)

2fyx:fxy(x,y),f12(x,y),Dxy(x,y),D12(x,y)

2fxy:fyx(x,y),f21(x,y),Dyx(x,y),D21(x,y)


This page titled 2.2: Partial Derivatives is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Michael Corral via source content that was edited to the style and standards of the LibreTexts platform.

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