core 3 differentiation learning objectives: review understanding of differentiation from core 1 and...
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Core 3 Differentiation
Learning Objectives:Learning Objectives:Review understanding of Review understanding of differentiation from Core 1 and 2differentiation from Core 1 and 2Understand how to differentiate eUnderstand how to differentiate exx
Understand how to differentiate ln Understand how to differentiate ln aaxx
Differentiation means……Differentiation means……
Finding the gradient function.Finding the gradient function.
The gradient function is used to calculate The gradient function is used to calculate
the gradient of a curve for any given the gradient of a curve for any given
value of x, so at any point.value of x, so at any point.
Differentiation Review
The Key Bit
The general rule (very important) is :-
If y = xn
dydx
= nxn-1
E.g. if y = x2
= 2xdydx
E.g. if y = x3
= 3x2dydx
E.g. if y = 5x4
= 5 x 4x3
= 20x3
dydxdydx
A differentiating Problem
The gradient of y = ax3 + 4x2 – 12x is 2 when x=1
What is a?dydx
= 3ax2 + 8x -12
When x=1dydx
= 3a + 8 – 12 = 2
3a - 4 = 23a = 6 a = 2
Finding Stationary Points
At a maximum At a minimum
dydx
=0 dydx
=0
+
dydx
> 0
+-
dydx
< 0
-
d2ydx2 < 0< 0
d2ydx2 > 0> 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
SummaryThe 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
Differentiation of ex
The gradient function f’(x )and the original The gradient function f’(x )and the original function f(x) are identical, therefore function f(x) are identical, therefore
The gradient function of eThe gradient function of ex x is eis exx
i.e. the derivative of ei.e. the derivative of exx is e is exx
If f(x) = ex f `(x) = ex
Also, if f(x) = aex f `(x) = aex
Differentiation of ex
Turn to page 90 and work through Turn to page 90 and work through Exercise AExercise A
Derivative of ln x
ln x is the inverse of eln x is the inverse of exx
The graph of y=ln x is a reflection of The graph of y=ln x is a reflection of y = ey = ex x in the line y = xin the line y = x
This helps us to differentiate ln xThis helps us to differentiate ln x
If y = ln x then If y = ln x then x = ex = ey y soso
x
1 So Derivative of ln x is
yedy
dx
xedx
dyy
11
dx
dy
dy
dx= 1
Differentiation of ln x
Live page
Summary - ln ax (1)f(x) = ln xf(x) = ln x
f’(1) = 1f’(1) = 1 the gradient at x=1 is 1
f’(4) = 0.25f’(4) = 0.25 the gradient at x=4 is 0.25
f(x) = ln 3xf(x) = ln 3x
f(x) = ln 17xf(x) = ln 17x
f’(1) = 1f’(1) = 1 the gradient at x=1 is 1
f’(4) = 0.25f’(4) = 0.25 the gradient at x=4 is 0.25
f’(1) = 1f’(1) = 1 the gradient at x=1 is 1
f’(4) = 0.25f’(4) = 0.25 the gradient at x=4 is 0.25
Summary - ln ax (2)For f(x) = ln For f(x) = ln axx
Whatever value a takes……
the gradient function is the same
f’(1) = 1f’(1) = 1 the gradient at x=1 is 1
f’(4) = 0.25f’(4) = 0.25 the gradient at x=4 is 0.25
f’(100) = 0.01f’(100) = 0.01f’(0.2) = 5f’(0.2) = 5
the gradient at x=100 is 0.01the gradient at x=0.2 is 5
The gradient is always the reciprocal of x
For f(x) = ln For f(x) = ln axx f `(x) = 1/xf `(x) = 1/x
ExamplesIf f(x) = ln If f(x) = ln 7xx f `(x) = 1/xf `(x) = 1/x
If f(x) = ln If f(x) = ln 11xx33
f(x) = ln f(x) = ln 11 ++ ln xln x33
Don’t know about ln ax3
f(x) = ln f(x) = ln 11 ++ 33 ln xln x
f `(x) = 3f `(x) = 3 (1/x)(1/x)
f `(x) = 3/xf `(x) = 3/x
Constants go in differentiation
If y = xn dydx
= nxn-1
if f(x) = aex f `(x) = aex
if g(x) = ln ax g`(x) = 1/x
Summary
if h(x) = ln axn h`(x) = n/xh(x) = ln a + n ln x
Differentiation of ex and ln x Classwork / HomeworkClasswork / Homework
Turn to page 92Turn to page 92
Exercise BExercise B
Q1 ,3, 5Q1 ,3, 5