# Core 3 Differentiation Learning Objectives:

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Core 3 Differentiation Learning Objectives:
Review understanding of differentiation from Core 1 and 2 Understand how to differentiate ex Understand how to differentiate ln ax

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.

If y = xn = nxn-1 The Key Bit dy dx
The general rule (very important) is :- If y = xn dy dx = nxn-1 E.g. if y = x2 = 2x dy dx E.g. if y = x3 = 3x2 dy dx E.g. if y = 5x4 = 5 x 4x3 = 20x3 dy dx

A differentiating Problem
The gradient of y = ax3 + 4x2 – 12x is 2 when x=1 What is a? dy dx = 3ax2 + 8x -12 When x=1 dy dx = 3a + 8 – 12 = 2 3a - 4 = 2 3a = 6 a = 2

Finding Stationary Points
At a maximum At a minimum + dy dx > 0 + d2y dx2 < 0 dy dx =0 - dy dx < 0 - d2y dx2 > 0 dy dx =0

Differentiation of ax Compare the graph of y = ax with the graph of its gradient function. Adjust the values of a until the graphs coincide.

Differentiation of ax Summary
The curve y = ax and its gradient function coincide when a = 2.718 The number 2.718….. is called e, and is a very important number in calculus See page 88 and 89 A1 and A2

Differentiation of ex

f `(x) = ex f `(x) = aex If f(x) = ex Also, if f(x) = aex
Differentiation of ex The gradient function f’(x )and the original function f(x) are identical, therefore The gradient function of ex is ex i.e. the derivative of ex is ex f `(x) = ex If f(x) = ex f `(x) = aex Also, if f(x) = aex

Differentiation of ex Turn to page 90 and work through Exercise A

Derivative of ln x = 1 ln x is the inverse of ex
The graph of y=ln x is a reflection of y = ex in the line y = x This helps us to differentiate ln x If y = ln x then x = ey so = 1 So Derivative of ln x is

Differentiation of ln x
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Differentiation of ln 3x
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Differentiation of ln 17x
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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 1 f’(4) = 0.25 the gradient at x=4 is 0.25 f(x) = ln 3x f’(1) = 1 the gradient at x=1 is 1 f’(4) = 0.25 the gradient at x=4 is 0.25 f(x) = ln 17x f’(1) = 1 the gradient at x=1 is 1 f’(4) = 0.25 the gradient at x=4 is 0.25

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 same f’(1) = 1 the gradient at x=1 is 1 f’(4) = 0.25 the gradient at x=4 is 0.25 f’(100) = 0.01 f’(0.2) = 5 the gradient at x=100 is 0.01 the gradient at x=0.2 is 5 The gradient is always the reciprocal of x For f(x) = ln ax f `(x) = 1/x

Examples f `(x) = 1/x If f(x) = ln 7x If f(x) = ln 11x3
Don’t know about ln ax3 f(x) = ln 11 + ln x3 f(x) = ln ln x f `(x) = 3 (1/x) Constants go in differentiation f `(x) = 3/x

= nxn-1 If y = xn f `(x) = aex if f(x) = aex if g(x) = ln ax
Summary dy dx = nxn-1 If y = xn f `(x) = aex if f(x) = aex if g(x) = ln ax g`(x) = 1/x if h(x) = ln axn h`(x) = n/x h(x) = ln a + n ln x

Differentiation of ex and ln x
Classwork / Homework Turn to page 92 Exercise B Q1 ,3, 5