2 x g x g dx d x f x f dx d x g x g x f dx d

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) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 x g x g dx d x f x f dx d x g x g x f dx d =
DIFFERENTIATION FORMULAS Example: If , then 3 2 3 x x y = 6 2 3 6 2 3 6 3 3 )] 2 3 ( [ 3 ) 3 )( 2 3 ( ) 3 ( ) ( ) 2 3 ( 2 3 ' x x x x x x x x x x d x x d x y = = = . ) 1 ( 6 ) 1 )( 2 ( 3 ) 2 2 ( 3 ) 2 3 ( 3 )] 2 3 ( [ 3 ' 4 6 2 6 3 2 6 2 3 3 6 2 3 x x x x x x x x x x x x x x x x y = = = = =
DIFFERENTIATION FORMULAS 7. Derivatives of Composition ( Chain Rule) Theorem: (The Chain Rule) If g is differentiable at x and if f is differentiable at g(x) , then the composition is differentiable at x. Moreover, if y=f(g(x)) and u = g(x) then y = f(u) and g f . dx du du dy dx dy =
DIFFERENTIATION FORMULAS Examples: Find y’. 1. 4 3 ) 1 ( 2 = x y . ) 1 ( 24 ) 3 ( ) 1 ( 8 ) 1 ( ) 1 )( 4 ( 2 ' 3 3 2 2 3 3 3 3 3 = = = x x x x x d x y 4 16 1 1 . 2 x y = 4 / 1 4 ) 16 1 ( 16 1 1 = = x x y . ) 16 1 ( 4 ) 16 ( ) 16 1 ( 4 1 ) 16 1 ( ) 16 1 ( 4 1 ' 4 / 5 4 / 5 4 / 5 x x x d x y = = =
IMPLICIT DIFEERENTIATION On occasions that a function F(x , y) = 0 can not be defined in the explicit form y = f(x) then the implicit form F ( x , y) = 0 can be used as basis in defining the derivative of y ( the dependent variable) with respect to x ( the independent variable). When differentiating F( x, y) = 0, consider that y is defined implicitly in terms of x , then apply the chain rule. As a rule, 1. Differentiate both sides of the equation with respect to x. 2. Collect all terms involving dy/dx on the left side of the equation and the rest of the terms on the other side. 3. Factor dy/dx out of the left member of the equation and solve for dy/dx by dividing the equation by the coefficient of dy/dx.
IMPLICIT DIFFERENTIATION Examples: Find y’. 1. 2. 1 ) ( 2 4 3 2 = y x y x ) ' 1 ( ) ( 8 ' 3 2 3 2 y y x y y x = ' ) ( 8 ) ( 8 ' 3 2 3 3 2 y y x y x y y x = ' ) ( 8 ) ( 8 ' 3 2 3 3 2 y y x y x y y x = 3 2 3 ) ( 8 3 ] ) ( 4 [ 2 ' y x y x y x y = 3 2 2 = y x y x ' 2 1 ) ( ) ( 2 2 yy x yd y d x = ' 2 1 ) 2 ( ' 2 yy x y y x = xy yy y x 2 1 ' 2 ' 2 = xy y x y 2 1 ] 2 [ ' 2 = . 2 2 1 ' 2 y x xy y =
HIGHER DERIVATIVES The notation dy/dx represent the first derivative of y with respect to x. And if dy/dx is differentiable, then the derivative of dy/dx with respect to x gives the second order derivative of y with respect to x and is denoted by . y wrt x of derivative n y general, In . . . . y wrt x of derivative third = ' ' y' y wrt x of derivative second ' ' y wrt x of derivative first ) ( ' ) ( 1 1 2 2 3 3 2 2 th n n n n n dx y d dx d dx y d dx y d dx d dx y d dx dy dx d dx y d y x f dx d dx dy y x f y = = = = = = = = d 2 y dx 2 = d dx dy dx
HIGHER DERIVATIVES Examples: 1. If then 2. if then , 1 3 2 3 5 = x x y ), 9 10 ( 9 10 ' 2 2 2 4 = = x x x x y ), 9 20 ( 2 18 40 " 2 3 = = x x x x y ). 9 60 ( 2 18 120 ' ' ' 2 2 = = x x y , 1 2 2 = y x , ' 0 ' 2 2 = yy x , ' y x y = . 1 ) ( ' ) 1 ( " 3 3 2 2 3 2 2 2 2 y y y x y x y y y x x y y xy y y = = = = =
Sample Problems Differentiate y with respect to x. Express dy/dx in simplest form. 1. y = x 2 5 7. y = 9 x 2 4 2. y = 5 x 3 6 7 x 8. y = x 5 x 2 2 5 3. y = 2 x 2 3 x 1 x 9. y = x 2 3 5 x 2 4. y = x 2 3 x 1 3 4 10. y = 2 2 x 5. y = x 1 4 x 3 11. x 2 y 2 = 9 6. y = 2 x 3 5 x 3 x 2 12. xy = x 2 y 1
Sample Problems Determine the derivative required: 1. y' of y = x 5 x 5 2. y''' of x 2 y 2 = 9 3.

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