Warm Up: h(x) is a composite function of f(x) and g(x). Find f(x) and g(x). 1. 2.

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Presentation transcript:

Warm Up: h(x) is a composite function of f(x) and g(x). Find f(x) and g(x)

THE CHAIN RULE Objective: To use the chain rule to find the derivative of composite functions.

What if we wanted to take the derivative of the following: 1.y=(2x+5) 2 2.y=(2x+5) 3 3.y=(2x+5) 10

If we had y= 6x -10, dy/dx = 6. EASY BREEZY What if we wrote y as a composite function? y = 2(3x -5) Outer function:Inner function: y = 2u u = 3x-5

The Chain Rule (used for composite functions) If y is a function of u, y= f(u), and u is a function of x, u=g(x), then y = f(u) = f(g(x)) and : If we have a composite function: Derivative of = derivative of X derivative of function outer function inner function (You may have composites within composites so may have to repeat)

If y=f (g(x)), then y’=f ’ (g(x)) ∙ g’(x) Derivative of outer function, times derivative of inner. Take a look: y=(2x-1) 2

Examples: Find the derivative. 1. y = (3x +4 ) 3 2.

Find the derivative. 1.y= sin(x 2 + x) 2.y= (x 3 +2x) -1 3.y= cot(x 2 )

Chain rule with products and quotients. Good times, good times…. 1.y = 3x(x 3 +2x 2 ) 3 2.y = 3.y = sin 2x cos2x

Quotient rule…do we need it?? You decide. Find the derivative:

Find the equation of the tangent line to the graph of at x = 3.

Determine the point(s) at which the graph of has a horizontal tangent.

Use the table of values to find the derivative. x f(x)2413 f ‘ (x) g(x)2341 g ‘ (x)2/73/74/75/7 FIND at x = 2.

Repeated Use of Chain Rule: Take derivative of outer function and work your way in until you have no more composites. f(g(h(x))) 1. f(x) = cos 2 (3x)

Find the derivative. 1.f(x) = tan(5 – sin2t) 2.y =