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4.3 Partial derivatives  (Page 3/11)

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If we choose to change y instead of x by the same incremental value h , then the secant line is parallel to the y -axis and so is the tangent line. Therefore, ∂ f / ∂ x represents the slope of the tangent line passing through the point ( x , y , f ( x , y ) ) parallel to the x -axis and ∂ f / ∂ y represents the slope of the tangent line passing through the point ( x , y , f ( x , y ) ) parallel to the y -axis . If we wish to find the slope of a tangent line passing through the same point in any other direction, then we need what are called directional derivatives , which we discuss in Directional Derivatives and the Gradient .

We now return to the idea of contour maps, which we introduced in Functions of Several Variables . We can use a contour map    to estimate partial derivatives of a function g ( x , y ) .

Partial derivatives from a contour map

Use a contour map to estimate ∂ g / ∂ x at the point ( 5 , 0 ) for the function g ( x , y ) = 9 − x 2 − y 2 .

The following graph represents a contour map for the function g ( x , y ) = 9 − x 2 − y 2 .

A series of concentric circles with the center the origin. The first is marked c = 0 and has radius 3; the second is marked c = 1 and has radius slightly less than 3; and the third is marked c = 2 and has radius slightly more than 2. The graph is marked with the equation g(x, y) = the square root of the quantity (9 – x2 – y2).
Contour map for the function g ( x , y ) = 9 − x 2 − y 2 , using c = 0 , 1 , 2 , and 3 ( c = 3 corresponds to the origin).

The inner circle on the contour map corresponds to c = 2 and the next circle out corresponds to c = 1 . The first circle is given by the equation 2 = 9 − x 2 − y 2 ; the second circle is given by the equation 1 = 9 − x 2 − y 2 . The first equation simplifies to x 2 + y 2 = 5 and the second equation simplifies to x 2 + y 2 = 8 . The x -intercept of the first circle is ( 5 , 0 ) and the x -intercept of the second circle is ( 2 2 , 0 ) . We can estimate the value of ∂ g / ∂ x evaluated at the point ( 5 , 0 ) using the slope formula:

∂ g ∂ x | ( x , y ) = ( 5 , 0 ) ≈ g ( 5 , 0 ) − g ( 2 2 , 0 ) 5 − 2 2 = 2 − 1 5 − 2 2 = 1 5 − 2 2 ≈ −1.688 .

To calculate the exact value of ∂ g / ∂ x evaluated at the point ( 5 , 0 ) , we start by finding ∂ g / ∂ x using the chain rule. First, we rewrite the function as g ( x , y ) = 9 − x 2 − y 2 = ( 9 − x 2 − y 2 ) 1 / 2 and then differentiate with respect to x while holding y constant:

∂ g ∂ x = 1 2 ( 9 − x 2 − y 2 ) −1 / 2 ( −2 x ) = − x 9 − x 2 − y 2 .

Next, we evaluate this expression using x = 5 and y = 0 :

∂ g ∂ x | ( x , y ) = ( 5 , 0 ) = − 5 9 − ( 5 ) 2 − ( 0 ) 2 = − 5 4 = − 5 2 ≈ −1.118 .

The estimate for the partial derivative corresponds to the slope of the secant line passing through the points ( 5 , 0 , g ( 5 , 0 ) ) and ( 2 2 , 0 , g ( 2 2 , 0 ) ) . It represents an approximation to the slope of the tangent line to the surface through the point ( 5 , 0 , g ( 5 , 0 ) ) , which is parallel to the x -axis .

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Use a contour map to estimate ∂ f / ∂ y at point ( 0 , 2 ) for the function

f ( x , y ) = x 2 − y 2 .

Compare this with the exact answer.

Using the curves corresponding to c = −2 and c = −3 , we obtain

∂ f ∂ y | ( x , y ) = ( 0 , 2 ) ≈ f ( 0 , 3 ) − f ( 0 , 2 ) 3 − 2 = −3 − ( −2 ) 3 − 2 · 3 + 2 3 + 2 = − 3 − 2 ≈ −3.146 .

The exact answer is

∂ f ∂ y | ( x , y ) = ( 0 , 2 ) = ( −2 y | ( x , y ) = ( 0 , 2 ) = −2 2 ≈ −2.828 .

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Functions of more than two variables

Suppose we have a function of three variables, such as w = f ( x , y , z ) . We can calculate partial derivatives of w with respect to any of the independent variables, simply as extensions of the definitions for partial derivatives of functions of two variables.

Definition

Let f ( x , y , z ) be a function of three variables. Then, the partial derivative of f with respect to x, written as ∂ f / ∂ x , or f x , is defined to be

∂ f ∂ x = lim h → 0 f ( x + h , y , z ) − f ( x , y , z ) h .

The partial derivative of f with respect to y , written as ∂ f / ∂ y , or f y , is defined to be

∂ f ∂ y = lim k → 0 f ( x , y + k , z ) − f ( x , y , z ) k .

The partial derivative of f with respect to z , written as ∂ f / ∂ z , or f z , is defined to be

∂ f ∂ z = lim m → 0 f ( x , y , z + m ) − f ( x , y , z ) m .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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