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The Derivative-Instantaneous rate of change

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camden
24 diapositives · 1.2K Views · Aug 08, 2014

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:. Find the derivative of. Related problems.

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  1. 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:

  2. Find the derivative of

  3. Related problems 1) Find the slope of f (x) at x = 3, x = -2 2) Write the equation of the tangent line at 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

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

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

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

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

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

  9. 3.3 Differentiation formulas Constant Simple Power rule Constant multiple rule Sum and difference rule

  10. Find the derivative function for: rewrite rewrite

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

  12. Differentiate Product rule

  13. Differentiate Quotient rule

  14. Find the derivative function for:

  15. 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.

  16. 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 = -16t2 +160t. 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? 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.

  17. A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t2 +160t. a) How high does the rock go? Maximum height occurs when v =0. s = -16t2 +160t v = s´= -32t + 160 -32t + 160 = 0 t = 5 sec. At t = 5, s = -16(5)2 +160(5) = 400 feet.

  18. A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t2 +160t. b) What is the velocity when the rock is 256 ft. above the ground on the way up? On the way down? -16t2 +160t = 256 -16t2 +160t –256=0 -16(t2 - 10t + 16)=0 -16(t – 2) (t- 8) = 0 t = 2 or t = 8 Set position = 256 Find the times v =-32t + 160 at t = 2 v=-32(2)+160 = 96 ft/sec. at t = 8 v=-32(8)+160 = -96 ft/sec Substitute the times into the velocity function

  19. A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec. Its height is given by s = -16t2 +160t. c) What is the acceleration of the rock at any time? s = -16t2 +160t v = s´= -32t + 160 a = v´ = s´´= -32ft/sec2 d) When does the rock hit the ground? At what velocity? s = -16t2 +160t = 0 t = 0 and t = 10 Set position = 0 v =-32t + 160 v = -32(10)+ 160 = -160 ft/sec.

  20. 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

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

  22. 3.5 Derivatives of Trigonometric Functions

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

  24. Find the derivatives