Partial Derivatives.pdf - Partial Derivatives Applied Mathematics 1 MAT538 Introduction In this section we will learn how to differentiate a function of

# Partial Derivatives.pdf - Partial Derivatives Applied...

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Partial Derivatives Applied Mathematics MAT538 1 Introduction In this section, we will learn how to differentiate a function of several independent variables. Let f ( x, y ) be the function of two independent variables, then the first-order partial derivatives or first partial derivatives with respect to x and y at point ( x 0 , y 0 ) can be denoted as ∂f ∂x ( x 0 , y 0 ) = f x ( x 0 , y 0 ) , and ∂f ∂y ( x 0 , y 0 ) = f y ( x 0 , y 0 ) . In partial derivative, when we differentiate the function with respect to any of independent variable, other independent variable(s) will be treated as constant(s). Example 1. Find the value of ∂f ∂x and ∂f ∂y at the point (4 , - 5) if f ( x, y ) = x 2 + 3 xy + y - 1. Solution ∂f ∂x = ∂x ( x 2 + 3 xy + y - 1 ) ; = 2 x + 3(1)( y ) + 0 - 0; = 2 x + 3 y. Thus ∂f ∂x (4 , - 5) = f x (4 , - 5) = 2(4) + 3( - 5) = - 7. ∂f ∂y = ∂y ( x 2 + 3 xy + y - 1 ) ; = 0 + 3( x )(1) + 1 - 0; = 3 x + 1 . Thus ∂f ∂y (4 , - 5) = f y (4 , - 5) = 3(4) + 1 = 13. Example 2. Find f y if f ( x, y ) = y sin( xy ) + ( x + y ) 6 + y y + 1 . Solution f y = ∂y [ y sin( xy )] + ∂y ( x + y ) 6 + ∂y y y + 1 ; = y ∂y [sin( xy )] + sin( xy ) ∂y [ y ] | {z } product rule + 6( x + y ) 5 ∂y ( x + y ) | {z } chain rule + ( y + 1) ∂y ( y ) - ( y ) ∂y ( y + 1) ( y + 1) 2 | {z } quotient rule ; = y cos( xy )( x ) + sin( xy )(1) + 6( x + y ) 5 (0 + 1) + ( y + 1)(1) - ( y )(1 + 0) ( y + 1) 2 ; = xy cos( xy ) + sin( xy ) + 6( x + y ) 5 + 1 ( y + 1) 2 . Practice Find all the first partial derivatives for the following functions 1. f ( x, y ) = 2 x 2 - 3 y - 4. ans f x = 4 x, f y = - 3 2. f ( x, y ) = x/y . ans f x = 1 y , f y = - x y 2 . 3. f ( x, y ) = p x 2 + y 2 . ans f x = x p x 2 + y 2 , f y = y p x 2 + y 2 . 4. f ( x, y ) = e xy ln( y ). ans f x = ye xy ln( y ) , f y = e xy y + xe xy ln( y ). 5. f ( x, y ) = 2 y - x x + y + 1 , ( - 1 , 1). ans f x = - 4 , f y = - 1. Page 1 of 6
Partial Derivatives Applied Mathematics MAT538 1.1 Implicit Differentiation Example 3. Let z = f ( x, y ), find ∂z ∂x and ∂z ∂y for the equation yz 2 - ln( z ) = x + y . Solution ∂x yz 2 - ∂x [ln( z )] = ∂x [ x ] + ∂x [ y ]; y (2 z ) ∂z ∂x - 1 z ∂z ∂x = 1 + 0; 2 yz - 1 z ∂z ∂x = 1; ∂z ∂x = 1 2 yz - 1 z . ∂y yz 2 - ∂y [ln( z )] = ∂y [ x ] + ∂y [ y ]; y (2 z ) ∂z ∂y + z 2 (1) - 1 z ∂z ∂y = 0 + 1; 2 yz - 1 z ∂z ∂y = 1 - z 2 ; ∂z ∂y = 1 - z 2 2 yz - 1 z . Alternatively, we can use Implicit Function Theorem to do the differentiation. The theorem state that if z = f ( x, y ) is differentiable and F ( x, y, z ) = 0 then z x = ∂z ∂x = - ∂F ∂x / ∂F ∂z = - F x F z , z y = ∂z ∂y = - ∂F ∂y / ∂F ∂z = - F y F z .

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