C1: Simple Differentiation
Learning Objective: to calculate
the gradient of a curve by applying
the rules of differentiation
The gradient of a curve
The gradient of a curve at a point is given by
the gradient of the tangent at that point.
Look at how the gradient changes as we move along a curve:
Differentiation
If we continued the process of differentiating from first
principles we would obtain the following results:
y
dy
dx
x
x2
x3
x4
x5
x6
1
2x
3x2
4x3
5x4
6x5
What pattern do you notice?
In general:
If y  x th e n
n
dy
 nx
n 1
dx
and when xn is preceded by a constant multiplier a we have:
n
If y = ax th e n
dy
dx
= anx
n 1
The gradient function
So the gradient of the tangent to the curve can be calculated
by differentiation.
dy
dx
represents the derivative of y with respect to x.
So if y =
x3
then:
dy
= 3x
2
dx
This notation can be adapted for other variables so, for
example:
ds
dt
represents the derivative of s with respect to t.
If the curve is written using function notation as y = f(x), then
the derived function can be written as f ′(x).
Task 1
Differentiate
1. y = x4
2. y = 3x4
3. y = 5x3
4. y = 2x8
5. y = -5x6
6. y = x12
7. y = -0.5x2
8. y = 2.5x4
9. y = 15x8
10. y = 2/3 x3
Differentiation
Find the gradient of the curve
y = 3x4 at the point (–2, 48).
Differentiating:
dy
= 12 x
3
dx
At the point (–2, 48) x = –2 so:
dy
= 1 2(  2 )
3
dx
= 12 × 8
= 96
The gradient of the curve y = 3x4 at the point (–2, 48) is –96.
Task 2
1. Find the gradient of the tangent of y = x4
at the point (3, 81).
2. Find the gradient of the curve whose
equation is y = 3x2 at the point (2, 12).
3. Find the gradients of the curve y = 2x2 at
the points C and D where the curve
meets the line y = x + 3.
You can apply the rules of differentiation to functions
involving negative or fractional powers and for polynomials.
Examples:
1. y = x-2
1. dy/dx = -2x-3
2. y = x1/2
2. dy/dx = ½ x-1/2
3. y = 3x2 + 5x - 6
3. dy/dx = 6x + 5
Task 3: Find dy/dx when y equals:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
2x2 -6x + 3
½ x2 + 12x
4x2 - 6
8x2 + 7x +12
5 + 4x – 5x2
x3 + x2 – x1/2
2x-3
1/3 x1/2 + 4x-2
5√x
x3(3x + 1)
Task 4:
1.
Find the gradient of the curve whose equation is
y = 2x2 – x – 1 at the point (2,5).
2.
Find the y co-ordinate and the value of the gradient at
the point P with x co-ordinate 1 on the curve with
equation y = 3 + 2x - x2.
3.
Find the co-ordinates of the point on the curve with
equation y = x2 + 5x – 4 where the gradient is 3.
4.
Find the point or points on the curve with equation f(x),
where the gradient is zero: a) f(x) = x3/2 – 6x + 1
b) f(x) = x-1 + 4x.
Second Order Derivatives
You can repeat the process of differentiation to
give a second order derivative.
The notation d2y/dx2 or f’’(x) is used.
Example:
y = 3x5 + 4/x2
y = 3x5 + 4x-2
dy/dx = 15x4 -8x-3
d2y/dx2 = 60x3 + 24x-4
= 60x3 + 24/x4
Task 5: Find dy/dx and d2y/dx2 when y equals:
1. 12x2 + 3x + 8
2. 15x + 6 +3/x
3. 9√x – 3/x2
4. (5x + 4)(3x – 2)
5. (3x+ 8)/(x2)
Task 6: Using different notation
1.
2.
3.
4.
5.
Find dθ/dt where θ = t2 – 3t
Find dA/dr where A = 2πr
Given that r = 12/t, find the value of dr/dt when t = 3.
The surface area, A cm2, of an expanding sphere of
radius r cm is given by A = 4πr2. Find the rate of
change of the area with respect to the radius at the
instant when the radius is 6cm.
The displacement, s metres, of a car from a fixed point
at time t seconds is given by s = t2 + 8t. Find the rate of
change of the displacement with respect to time at the
instant when t = 5.