Presentation on theme: "Core 3 Differentiation Learning Objectives:"— Presentation transcript:
1 Core 3 Differentiation Learning Objectives: Review understanding of differentiation from Core 1 and 2Understand how to differentiate exUnderstand how to differentiate ln ax
2 Differentiation Review Differentiation means……Finding the gradient function.The gradient function is used to calculate the gradient of a curve for any given value of x, so at any point.
3 If y = xn = nxn-1 The Key Bit dy dx The general rule (very important) is :-If y = xndydx= nxn-1E.g. if y = x2= 2xdydxE.g. if y = x3= 3x2dydxE.g. if y = 5x4= 5 x 4x3= 20x3dydx
4 A differentiating Problem The gradient of y = ax3 + 4x2 – 12x is 2 when x=1What is a?dydx= 3ax2 + 8x -12When x=1dydx= 3a + 8 – 12 = 23a - 4 = 23a = 6a = 2
5 Finding Stationary Points At a maximumAt a minimum+dydx> 0+d2ydx2< 0dydx=0-dydx< 0-d2ydx2> 0dydx=0
6 Differentiation of axCompare the graph of y = ax with the graph of its gradient function.Adjust the values of a until the graphs coincide.
7 Differentiation of ax Summary The curve y = ax and its gradient function coincide when a = 2.718The number 2.718….. is called e, and is a very important number in calculusSee page 88 and 89 A1 and A2
9 f `(x) = ex f `(x) = aex If f(x) = ex Also, if f(x) = aex Differentiation of exThe gradient function f’(x )and the original function f(x) are identical, thereforeThe gradient function of ex is exi.e. the derivative of ex is exf `(x) = exIf f(x) = exf `(x) = aexAlso, if f(x) = aex
10 Differentiation of exTurn to page 90 and work through Exercise A
11 Derivative of ln x = 1 ln x is the inverse of ex The graph of y=ln x is a reflection ofy = ex in the line y = xThis helps us to differentiate ln xIf y = ln x then x = ey so= 1So Derivative of ln x is
15 Summary - ln ax (1) f(x) = ln x f’(1) = 1 f’(4) = 0.25 f(x) = ln 3x the gradient at x=1 is 1f’(4) = 0.25the gradient at x=4 is 0.25f(x) = ln 3xf’(1) = 1the gradient at x=1 is 1f’(4) = 0.25the gradient at x=4 is 0.25f(x) = ln 17xf’(1) = 1the gradient at x=1 is 1f’(4) = 0.25the gradient at x=4 is 0.25
16 Summary - ln ax (2) For f(x) = ln ax For f(x) = ln ax f `(x) = 1/x Whatever value a takes……the gradient function is the samef’(1) = 1the gradient at x=1 is 1f’(4) = 0.25the gradient at x=4 is 0.25f’(100) = 0.01f’(0.2) = 5the gradient at x=100 is 0.01the gradient at x=0.2 is 5The gradient is always the reciprocal of xFor f(x) = ln axf `(x) = 1/x
17 Examples f `(x) = 1/x If f(x) = ln 7x If f(x) = ln 11x3 Don’t know about ln ax3f(x) = ln 11 + ln x3f(x) = ln ln xf `(x) = 3 (1/x)Constants go in differentiationf `(x) = 3/x
18 = nxn-1 If y = xn f `(x) = aex if f(x) = aex if g(x) = ln ax Summarydydx= nxn-1If y = xnf `(x) = aexif f(x) = aexif g(x) = ln axg`(x) = 1/xif h(x) = ln axnh`(x) = n/xh(x) = ln a + n ln x
19 Differentiation of ex and ln x Classwork / HomeworkTurn to page 92Exercise BQ1 ,3, 5