Differentiation
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Differentiation by first principles
Differentiation of polynomials
Examples of derivative functions & gradients
Graphs of derivative functions
Tangents
Normal lines
Finding stationary points
Anti-differentiation
Finding the original function
VCE Maths Methods - Unit 2 - Differentiation
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Differentiation by first principles
• As the average rate is calculated over smaller changes in x (∆x = h), it
approaches the instantaneous gradient.
y = x2
m=
Gradient at x = 1
(2,4)
4 −1 3
= =3
2−1 1
2.25−1 1.25
m=
=
=2.5
1.5−1 0.5
3 9
( , )
2 4
3
1.21−1 0.21
m=
=
=2.1
1.1−1 0.1
1.25
m=
(1,1)
h = 0.5 h =1
VCE Maths Methods - Unit 2 - Differentiation
1.0201−1 0.0201
=
=2.01
1.01−1
0.01
2
Differentiation by first principles
• The instantaneous gradient can be found from the limit function:
y2
y1
f (x +h)−f (x)
limh→0
h
∆x
• The gradient of the curve y = x2:
f (x +h)−f (x)
limh→0
h
x 2 +2hx +h 2 −x 2
limh→0
h
(x +h)2 −(x)2
limh→0
h
2hx +h 2
limh→0
h
limh→0 2x +h =2x
The gradient at any point on the curve y = x2 is 2x
VCE Maths Methods - Unit 2 - Differentiation
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Differentiation of polynomials
• First principles can be used to find an overall rule for differentiating any
polynomial function.
f (x) = x 2 →f '(x) =2x
d
f (x) =f '(x)
dx
d n
x =nx n−1
dx
f (x) = x 5 →f '(x) =5x 4
Rules for differentiation:
d n
ax = anx n−1 Common factors / constants can be multiplied into the derivative.
dx
f (x) =2x 3 →f '(x) =2×3x 2 = 6x 2
d
c =0
dx
The derivative of a constant is zero.
f (x) =2 →f '(x) = 0
If f(x)=g(x)+h(x) then f'(x)=g'(x)+h'(x) The derivative of a sum of functions is the
sum of the derivatives.
f (x) = 3x 3 −4x +6 →f '(x) = 9x 2 −4
VCE Maths Methods - Unit 2 - Differentiation
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Graphs of derivative functions
y = x2 + 2
gradient = +4
gradient = - 4
y = 2x
gradient = 0
y=4
y=0
y=-4
VCE Maths Methods - Unit 2 - Differentiation
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Examples of derivative functions & gradients
Find the derivative of the function:
3
2
f (x) =2x +3x +4x +2
f '(x) =(3×2x 2 )+(2×3x)+4
f '(x) = 6x 2 +6x +4
Find the derivative of the function:
1
2
1
f (x) =5 x + 2 f (x) =5x +x −2
x 1
5 −2
f '(x) = x −2x −3
2
5
2
f '(x) =
− 3
2 x x
Find the gradient of the curve at the
point where x = 2 :
Find the gradient of the curve at the
point where x = 3 :
f (x) = 4x 2 −3x
f '(x) = 8x −3
f (x) =2x 2 −12x +10
f '(2) = 8×2−3
f '(2) =13
f '(3) = 4×3−12
VCE Maths Methods - Unit 2 - Differentiation
f '(x) = 4x −12
f '(3) = 0
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Tangent lines
• The tangent line touches the curve at one point.
• It has the same gradient as the curve at the point of contact.
x2
y = − +2x
2
Tangent line
3 15
( , )
2 8
Find equation:
Find gradient:
x2
f (x) = − +2x
2
f '(x) = −x +2
3
3
f '( ) = − +2
2
2
y − y 1 =m(x −x1 )
15 1
3
y−
= (x − )
8 2
2
1 3 15
y = x− +
2 4 8
1 9
y = x+
2 8
3 1
f '( ) =
2 2
VCE Maths Methods - Unit 2 - Differentiation
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Normal lines
• The normal line is at right angles to the tangent & through the same point.
• The gradient is the opposite sign & reciprocal of the tangent.
x2
y = − +2x
2
Tangent line
Normal line
3 15
( , )
2 8
Find gradient:
3 1
f '( ) =
2 2
m = −2
VCE Maths Methods - Unit 2 - Differentiation
Find equation:
y − y 1 =m(x −x1 )
15
3
y−
= −2(x − )
8
2
15
y = −2x +3+
8
39
y = −2x +
8
8
Finding stationary points
• Differentiation can be used to find the local maximum or minimum points on
a graph.
• At the stationary points, the gradient is equal to 0.
3
y = x −27x
y =(−3)3 −(27×−3) =54
Local maximum
Find where
the gradient is zero:
y = x 3 −27x
x = −3
dy
= 3x 2 −27
dx
0 = 3x 2 −27
27 = 3x 2
9 = x2
x =3
Local minimum
x =±3
y = 33 −(27×3) = −54
VCE Maths Methods - Unit 2 - Differentiation
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Anti-differentiation
• The anti-derivative of a function can be found by the reverse process of
differentiation.
• Anti-differentiation is used to find the total from a rate.
n+1
∫ f'(x) dx = f(x) + c
3
x
f (x) = x →F (x) = +c
3
2
∫ f(x) dx = F(x) + c
x
∫ x dx = n +1+c (n ≠ -1)
6
x
f (x) = x 5 →F (x) = +c
6
n
f (x) = x
3
2
2x
→F (x) =
+c
3
1
2
Rules for anti-differentiation:
Common factors / constants can be
multiplied into the anti-derivative.
f (x) =5x 3
x3
5x 3
→F (x) =5× +c =
+c
3
3
The anti-derivative of a sum of functions If f(x)=g(x)+h(x) then F(x)=G(x)+H(x)+c
is the sum of the anti-derivatives.
3 4
2
f (x) = 3x 3 −4x +6 →F (x) =
VCE Maths Methods - Unit 2 - Differentiation
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x −2x +6x +c
10
Examples of derivative functions & gradients
Find an anti-derivative of the function:
f (x) =2x 3 +3x 2 +4x +2
2x 4 3x 3 4x 2
F (x) =
+
+
+2x +c
4
3
2
x4 3
F (x) = +x +2x 2 +2x +c
2
VCE Maths Methods - Unit 2 - Differentiation
Find the anti-derivative of the function:
1
1
f (x) =5 x + 2 f (x) =5x 2 +x −2
x
3
5 2 x −1
F (x) =
x +
+c
3
−1
 
2
10 1
F (x) =
− +c
3 x3 x
11
Finding the original function
• Find the equation of the curve passing through the point (2, −2), with a
gradient function 2x + 3.
f (x) =2x +3
2x 2
F (x) =
+3x +c
2
F (x) = x 2 +3x +c
(2,−2)
y = x 2 +3x +c
2
−2= (2) +3 (2) +c
−2= 4+6+c
c = −12
VCE Maths Methods - Unit 2 - Differentiation
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