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The Derivative-Instantaneous rate of change The derivative of a function, f at a specific value of x, say a is a value given by: The derivative of a function, f as a function of x, is called f (x) and is given by:

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Find the derivative of

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Related problems 1) Find the slope of f (x) at x = 3, x = -2 3) Find the point on f (x) for which the slope is 2 4) Find the point for which f (x) has a horizontal tangent line 2) Write the equation of the tangent line at x = -2

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Solutions 1) Find the slope of f (x) at x = 3, x = -2 2) Write the equation of the tangent line at x = -2 so use the point-slope formula Find the value of y

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Solutions 3) Find the point on f (x) for which the slope is 2 4) Find the point for which f (x) has a horizontal tangent line The point is (1, 6) The point is (2/3, 17/3)

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Figure 2.7: Derivatives at endpoints are one-sided limits. Derivatives at Endpoints are one-sided limits.

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How a derivative can fail to exist A corner A vertical tangent A discontinuity Which of the three examples are the functions continuous?

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The graph of a function The graph of the derivative (slope) of the function Where f(x) is increasing (slope is positive) Where f(x) is decreasing (Slope is negative) Horizontal tangent (slope =0)

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3.3 Differentiation formulas Simple Power rule Sum and difference rule Constant multiple rule Constant

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Find the derivative function for: rewrite

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Rules for Finding Derivatives u and v are functions of x. Simple Power rule Sum and difference rule Constant multiple rule Product rule Quotient rule

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Differentiate Product rule

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Differentiate Quotient rule

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Find the derivative function for:

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Velocity. The particle is moving forward for the first 3 seconds and backwards the next 2 sec, stands still for a second and then moves forward. forward motion means velocity is positive backward motion means velocity is negative If velocity = 0, object is standing still.

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The graphs of s and v as functions of time; s is largest when v = ds/dt = 0. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock’s velocity graphed here as a straight line. a) How high does the rock go? b) What is the velocity when the rock is 256 ft. above the ground on the way up? On the way down? c) What is the acceleration of the rock at any time? d) When does the rock hit the ground? At what velocity? 3.4 applications A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t 2 +160t.

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a) How high does the rock go? Maximum height occurs when v =0. -32t + 160 = 0 v = s´= -32t + 160 t = 5 sec. s = -16t 2 +160t At t = 5, s = -16(5) 2 +160(5) = 400 feet.

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A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t 2 +160t. b) What is the velocity when the rock is 256 ft. above the ground on the way up? On the way down? v =-32t + 160 at t = 2 v=-32(2)+160 = 96 ft/sec. at t = 8 v=-32(8)+160 = -96 ft/sec -16t 2 +160t = 256 -16t 2 +160t –256=0 -16(t 2 – 10t + 16)=0 -16(t – 2) (t- 8) = 0 t = 2 or t = 8 Find the times Substitute the times into the velocity function Set position = 256

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A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t 2 +160t. c) What is the acceleration of the rock at any time? d) When does the rock hit the ground? At what velocity? s = -16t 2 +160t v = s´= -32t + 160 a = v´ = s´´= -32ft/sec 2 s = -16t 2 +160t = 0 t = 0 and t = 10 v =-32t + 160 v = -32(10)+ 160 = -160 ft/sec. Set position = 0

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3.5 Derivatives of trig functions-formulas needed sin(x+h) = sin x*cos h+cos x*sin h cos(x+h) = cos x*cos h- sin x*sin h

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Derivative of y = sin x 0 +cos(x)*1 = cos (x)

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3.5 Derivatives of Trigonometric Functions

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Figure 25: The curve y´ = –sin x as the graph of the slopes of the tangents to the curve y = cos x. Slope of y = cos x

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Find the derivatives

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