1. 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

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 = 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

4. 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

5. 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

6. 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.

7. 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

8. Differentiation of ex

9. 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

10. Differentiation of ex
Turn 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 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

12. Differentiation of ln x
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13. Differentiation of ln 3x
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14. Differentiation of ln 17x
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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 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

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 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

17. 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

18. = 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

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