For any function f ( x,y ), the first partial derivatives are represented by  f  f

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# For any function f ( x,y ), the first partial derivatives are represented by  f  f - PowerPoint PPT Presentation

For any function f ( x,y ), the first partial derivatives are represented by  f  f — = f x and — = f y  x  y. For example, if f ( x,y ) = log( x sin y ), the first partial derivatives are  f  f — = f x = and — = f y =  x  y. 1 — x. cos y

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ff

— = fxand — = fy

xy

For example, if f(x,y) = log(x sin y), the first partial derivatives are

ff

— = fx = and — = fy =

xy

1

x

cos y

—— = cot y

sin y

If a function f from Rn to R1 has continuous partial derivatives, we say that f belongs to class C1. We can see that f(x,y) = log(x sin y) belongs to class C1 when its domain is defined so that

x sin y > 0.

If each of the partial derivatives of f belongs to class C1, then we say that

f belongs to class C2.

f 1 f

f(x,y) = log(x sin y), — = fx = —and — = fy = cot y

xxy

We can calculate higher order (and mixed) partial derivatives:

 f

— ( — ) = — ( fx) = ( fx)x = fxx =

x xx

1

– —

x2

 f

— ( — ) = — ( fy) = ( fy)y = fyy =

y yy

1

– —— = – csc2y

sin2y

 f

— ( — ) = — ( fx) = ( fx)y = fxy =

y xy

0

 f

— ( — ) = — ( fy) = ( fy)x = fyx =

x yx

0

Let f(x,y) = sin(xy)

fx = fy =

fxx = fyy =

fxy =

fyx =

y cos(xy)

x cos(xy)

– y2 sin(xy)

– x2 sin(xy)

cos(xy) – xy sin(xy)

cos(xy) – xy sin(xy)

f(x+x , y) – f(x , y)

—————————

x

(x0 , y0+y)

(x0+x , y0+y)

fx(x , y) 

f(x , y+y) – f(x , y)

—————————

y

fy(x , y) 

Consider

fxy(x0 , y0) 

(x0+x , y0)

(x0 , y0)

fx(x0 , y0+y) – fx(x0 , y0)

——————————

y

f(x0+x , y0) – f(x0 , y0)

Substitute ————————— in place of fx(x0 , y0) , and

x

f(x0+x , y0+y) – f(x0 , y0+y)

substitute ————————————— in place of fx(x0 , y0+y) .

x

Now consider

fy(x0+x , y0) – fy(x0 , y0)

fyx(x0 , y0)  ——————————

x

f(x0 , y0+y) – f(x0 , y0)

Substitute ————————— in place of fy(x0 , y0) , and

y

f(x0+x , y0+y) – f(x0+x, y0)

substitute ————————————— in place of fy(x0+x , y0) .

y

Note that the results are the same in both cases suggesting that

fxy = fyx .

Look at Theorem 1 on page 183 (and note how this result can be extended to partial derivatives of any order).