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Unit 3 Outcome 2

Unit 3 Outcome 2. Further Calculus. Reminder on Derivative Graphs. The graph of y = g(x) is given below. y =g(x). X. 5 6 7 8 9 10. x. 6. 8. SPs occur at x = 6 & x = 8. g'(x). + 0 - 0 +. New y-values. The graph of y = g '(x) is.

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Unit 3 Outcome 2

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  1. Unit 3 Outcome 2 Further Calculus

  2. Reminder on Derivative Graphs The graph of y = g(x) is given below y =g(x) X 5 6 7 8 9 10 x 6 8 SPs occur at x = 6 & x = 8 g'(x) + 0 - 0 + New y-values

  3. The graph of y = g '(x) is y = g '(x) X 6 8

  4. Derivative Graph of y = sinx y = sinx -/2 3/2 /2 5/2 SPs at x = -/2 , x = /2 , x = 3/2 & x =5/2 x -/2 /2  3/2  5/2 f (x) 0 + 0 - 0 + 0 New y-values

  5. /2 3/2 -/2 2 5/2 Roller-coaster from 0 to 2 ie shape of y = cosx Hence If y = sinx thendy/dx = cosx

  6. A similar argument shows that …. If y = cosx thendy/dx = -sinx If we keep differentiating the basic trig functions we get the following cycle f(x) f  (x) sinx cosx cosx -sinx -sinx -cosx -cosx sinx

  7. Example If g(x) = cosx – sinx then find g  (/2) . ********** g  (x) = -sinx – cosx g  (/2) = -sin/2 - cos/2 = -1 - 0 = -1 Example Prove that y = 7x – 2cosx is always increasing! ********* dy/dx = 7 + 2sinx since -2 < 2sinx < 2 then 5 < dy/dx < 9 Gradient is always positive so function is always increasing.

  8. Example Find the equation of the tangent to y = sinx – cosx at the point where x = /4 . ******** Point of contact If x = /4 then y = sin/4 – cos/4 = 1/2 - 1/2 = 0 (/4 ,0) Gradient dy/dx = cosx + sinx when x = /4 then dy/dx = cos/4 + sin/4 = 1/2 + 1/2 = 2/2 = 2 Line Using y – b = m(x – a) we get y - 0 = 2(x - /4) ie y = 2x - 1/4 2

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