Differentiation
Calculus 2.7
Finding the Gradient Function (Differentiation)
Step 1: Choose or
Step 2: Multiply exponent by coefficient
Step 3: Subtract 1 from the exponent
f (x) = 4x3
f ‘ (x) = 12x2
y = x4 - 2x
= 4x3 - 2x0 = 4x3 - 2
f (x) = 8x + 5
f ‘ (x) = 8 + 0 = 8
f (x) = 0.2x4 + 2x
f ‘ (x) = 0.8x3 + 2
Using a differentiated polynomial to calculate the gradient (p 19)
Calculate the gradient of f(x) = x3 where x = 1
Step 1: Differentiate the function
f’(x) = 3x2
Step 2: Substitute x = 1 into the derivative
f’(1) = 3(1)2
f’(1) = 3(1)
f’(1) = 3
So the gradient of f(x) = x3 when x = 1 is 3
Using a differentiated polynomial to calculate the gradient (p 18)
Calculate the gradient of y = 3x2 + 4x - 5 where x = 5
Step 1: Differentiate the function
dy/dx = 6x + 4
Step 2: Substitute x = 5 into the derivative
Gradient = 6(5) + 4
Gradient = 34
So the gradient of y = 3x2 + 4x - 5 where x = 5 is 34
Using a differentiated polynomial to calculate the gradient (p 18)
Calculate the gradient of y = -3x6 - 42x3 where x = 0.18
Step 1: Differentiate the function
dy/dx = -18x5 - 126x2
Step 2: Substitute x = 0.18 into the derivative
Gradient = -18(0.18)5 - 126(0.18)2
Gradient = -4.09 (2 d.p.)
So the gradient of y = -3x6 - 42x3 where x = 0.18 is -4.09
Using a differentiated polynomial to calculate the gradient (p19 q5)
Calculate the gradient of f(x) = 5x where x = 3
Step 1: Differentiate the function
f’(x) = 5
Step 2: Substitute x = 3 into the derivative
This is a straight line with a gradient of 5. The gradient is 5 for all values of x.
Using a differentiated polynomial to calculate the gradient (similar to p19 q8)
Calculate the gradient of
y = (x + 2)(3 + x) at the point (-1, 2)
Step 1: Identify our x value
x = -1
Step 2: Expand and simplify the brackets
y = 3x + x2 + 6 + 2x = x2 + 5x + 6
Step 3: Differentiate the function
dy/dx = 2x + 5
Step 3: Substitute x = -1 into the derivative
Gradient = 2(-1) = -2
Using a differentiated polynomial to calculate the gradient (p19 q6)
Calculate the gradient of y = x(7 - 2x) where x = -2
Step 1: Expand and simplify the brackets
y = 7x - 2x2
Step 2: Differentiate the function
dy/dx = 7 - 4x
Step 3: Substitute x = -2 into the derivative
Gradient = 7 - 4(-2)
Gradient = 7 - (-8)
Gradient = 15
Using a differentiated polynomial to calculate the gradient (p19 q7)
Calculate the gradient of where the curve crosses the y-axis
Step 1: Identify x
x = 0, because this is the y-axis
Step 2: Convert to decimals or fractions
f(x) = 0.5x3 - 0.5x or
Step 3: Differentiate the function
f’(x) = 1.5x2 - 0.5 or
Step 4: Substitute x = 0 into the derivative
f’(0) = 1.5(0)2 - 0.5 =-0.5
Using a differentiated polynomial to calculate the gradient (p19 q9)
Step 1: Convert to fractions
Step 2: Differentiate and substitute x = 9
Given the equation and a value for y,
calculate the gradient (p20 - 21)
Calculate the gradient of f(x) = -x2 + 4 where f(x) = 3
Step 1: Solve to find x value(s)
3 = -x2 + 4
-1 = -x2
1 = x2
x = ±1
Step 2: Differentiate the function
f’(x) = -2x
Step 3: Substitute x values into the derivative
f’(1) = -2(1) = -2 and f’(-1) = -2(-1) = 2
Gradient is -2 at (1, 3) and gradient is 2 at (-1, 3)
Given the equation and a value for y,
calculate the gradient (p21 q2)
Calculate the gradient of y = x3 where y = 8
Step 1: Solve to find x value(s)
8 = x3
x = 2
Step 2: Differentiate the function
= 3x2
Step 3: Substitute x values into the derivative
Gradient = 3(2)2
Gradient = 12
Gradient is 12 at (2, 8)
Given the equation and a value for y,
calculate the gradient (p21 q4)
Calculate the gradient of f(x) = x2 - 9 where f(x) = 0
Step 1: Solve to find x value(s)
0 = x2 - 9
9 = x2
x = ±3
Step 2: Differentiate the function
f’(x) = 2x
Step 3: Substitute x values into the derivative
f’(3) = 2(3) = 6
f’(-3) = 2(-3) = -6
Gradient is 6 when (3, 0) and gradient is -6 when (-3, 0)
Given the equation and a value for y,
calculate the gradient (p21 q5)
Calculate the gradient of y = x(5 - x) where y = 0
Step 1: Solve to find x value(s)
0 = x(5 - x)
x = 0
5 - x = 0 → x = 5
Step 2: Differentiate the function
y = 5x - x2
= 5 - 2x
Step 3: Substitute x values into the derivative
Gradient = 5 - 2(5) = -5
Gradient = 5 - 2(0) = 5
Gradient is -5 at (5, 0) and gradient is 5 at (0, 0)
Given the equation and coordinates of a point, find the equation of a tangent (p22 - 23)
Find the equation of the tangent to the curve
y = 3x2 - 2x + 4 at the point (1, 5)
Step 1: Differentiate the function
= 6x -2
Step 2: Find the gradient for the given x value
Gradient = 6(1) - 2 = 4
Step 3: Use y = mx + c to find out the tangent equation
y = mx + c
5 = (4)(1) + c
c = 1
So y = 4x + 1 is the equation of the tangent
Given the equation and coordinates of a point, find the equation of a tangent (p22 - 23)
Find the equation of the tangent to the curve
f(x) = x3 - 2x at the point where x = 2
Step 1: Find the y value by substituting x into equation
f(2) = (2)3 - 2(2) = 8 - 4 = 4 → So y = 4
Step 2: Differentiate the function
f’(x) = 3x2 - 2
Step 3: Find the gradient for the given x value
f’(2) = 3(2)2 - 2 = 10 → m = 10
Step 4: Use y = mx + c to find out the tangent equation
(4) = (10)(2) + c → c = -16
So y = 10x - 16 is the equation of the tangent
Given the equation and coordinates of a point, find the equation of a tangent (p23 q2)
Find the equation of the tangent to the curve
f(x) = 2x3 + 7x at the point (1, 9)
Step 1: Differentiate the function
f’(x) = 6x2 + 7
Step 2: Find the gradient for the given x value
f’(1) = 6(1)2 + 7 = 13
Step 3: Use y = mx + c to find out the tangent equation
y = mx + c
(9) = (13)(1) + c
c = 9 - 13 = -4
So y = 13x - 4 is the equation of the tangent
Given the equation and coordinates of a point, find the equation of a tangent (p23 q4)
Find the equation of the tangent to the curve
f(x) = 5 - 3x + x2 at the point (-1, 9)
Step 1: Differentiate the function
f’(x) = -3 + 2x
Step 2: Find the gradient for the given x value
f’(-1) = -3 + 2(-1) = -5
Step 3: Use y = mx + c to find out the tangent equation
y = mx + c
(9) = (-5)(-1) + c
c = 4
So y = -5x + 4 is the equation of the tangent
Given the equation and coordinates of a point, find the equation of a tangent (p23 q4)
Find the equation of the tangent to the curve
y = (x + 3)(x - 2) where x = 3
Step 1: Find the y value
y = x2 + x - 6
y = (3)2 + (3) - 6 = 9 + 3 - 6 = 6 → So y = 6
Step 1: Differentiate the function
dy/dx = 2x + 1
Step 2: Find the gradient for the given x value
Gradient = 2(3) + 1 = 7
Step 3: Use y = mx + c to find out the tangent equation
(6) = (7)(3) + c → c = -15
So y = 7x - 15 is the equation of the tangent
Given the equation and coordinates of a point, find the equation of a tangent (p23 q7)
Find the equation of the tangent to the curve y = 6x2 - 24 where y = 0
Step 1: Find the x value(s)
0 = 6x2 - 24
6x2 = 24
x2 = 4
x = ±2 → x1 = 2, x2 = -2
Step 2: Differentiate the function
dy/dx = 12x
Step 3: Find the gradients for the given x values
m1 = 12(2) = 24
m2 = 12(-2) = -24
Step 3: Use y = mx + c to find out the tangent equation
Equation 1: (0) = (24)(2) + c → c = -48
So y = 24x - 48 is the equation of the tangent at (2, 0)
Equation 2: (0) = (-24)(-2) + c → c = -48
So y = -24x - 48 is the equation of the tangent at (-2, 0)
Using a differentiated polynomial to locate points where the curve has a given gradient (p25 q1)
Locate the points where y = x2 has the gradient m = 6
Step 1: Differentiate the function
dy/dx = 2x
Step 2: Make the derivative equal to the given gradient and solve for x
6 = 2x
x = 3
Step 3: Substitute x value to find y value
y = (3)2
y = 9
So for the curve y = x2 the gradient is 6 at the point (3, 9)
Video of previous slide
Using a differentiated polynomial to locate points where the curve has a given gradient (p25 q7)
Locate the points where h(x) = 11 + x - ⅓x3 has a gradient of 0
Step 1: Differentiate the function
h’(x) = 1 - x2
Step 2: Make the derivative equal to the given gradient and solve for x
0 = 1 - x2
x2 = 1
x = ± 1
Step 3: Substitute x values to find y value
h(1) = 11 + (1) - ⅓(1)3 = 11.67 (2 d.p.)
h(-1) = 11 +(-1) - ⅓(-1)3 = 10.33 (2 d.p.)
So for the curve h(x) = 11 + x - 1/3x3 the gradient is 0 at the points
(1, 11.67) and (-1, 10.33)
Putting it all together (P26 Q1)
Find the coordinates of a point on the curve y = 11 - 3x + 0.5x2 where the gradient is -4
Step 1: Find the derivative of the function
dy/dx = -3 + x
Step 2: Make the derivative equal to the gradient and solve for x
-4 = -3 + x
x = -1
Step 3: Substitute x back into the function to find y
y = 11 - 3(-1) + 0.5(-1)2
y = 11 + 3 + 0.5 = 14.5
Coordinate where gradient is -4 is (-1, 14.5)
Putting it all together (P26 Q4)
Find the gradient of the curve g(x) = 2.5x2 - 1.5x where the graph crosses the y-axis
Step 1: Differentiate the function
g’(x) = 5x - 1.5
Step 2: Substitute x value into derivative
We know that x = 0 when the graph crosses the y-axis
g’(0) = 5(0) - 1.5
g’(0) = -1.5
So the gradient is -1.5 when then the graph crosses the y-axis
(This is the same as m = -3/2)
Locating turning points and
determining their nature (p28 - 30)
The turning points are where the curve ‘turns’. At these points the slope is flat, so the gradients is 0.
Turning point
m = 0
Locating turning points and
determining their nature (p28 -30)
The ‘nature’ of the turning point can be a maximum or a minimum
This turning point is a minimum
Locating turning points and
determining their nature (P30 Q1)
Locate the turning points and determine the nature of
y = x2 - 8x + 3
Step 1: Find derivative of the function
dy/dx = 2x - 8
Step 2: Make the derivative equal to 0 and solve for x
0 = 2x - 8
8 = 2x
x = 4
Step 3: Find y
y = (4)2 - 8(4) + 3
y = 16 - 32 + 3 = -13
Step 4: Write the coordinate of turning point(s)
Turning point is at (4, -13)
Step 5: Find second derivative
d2y/dx2 = 2
Step 6: Determine nature of turning point
Because second derivative is positive, the turning point is a minimum
Locating turning points and
determining their nature (P30 Q5)
Locate the turning points and determine the nature of
y = -x3 - 6x2 + 15x - 1
Step 1: Find derivative of the function
dy/dx = -3x2 - 12x + 15
Step 2: Make the derivative equal to 0 and solve for x
0 = -3x2 - 12x + 15
0 = x2 + 4x - 5 (Divide by -3)
0 = (x + 5)(x - 1)
x1 = -5 and x2 = 1
Step 3: Find y
y1 = -(-5)3 - 6(-5)2 + 15(-5) - 1 = -101
y2 = -(1)3 - 6(1)2 + 15(1) - 1 = 7
Step 4: Write the coordinate of turning point(s)
(-5, -101) and (7, 1) are the turning points
Step 5: Find second derivative
d2y/dx2 = -6x - 12
Step 6: Determine nature of turning point
-6(-5) - 12 = 18, so the point (-5, -101) is a minimum
-6(1) - 12 = -18, so the point (7, 1) is a maximum
Finding where functions are increasing and decreasing (P32 Q1)
x = 4
Determine the x values for which y = -x2 + 8x + 3 is decreasing.
Step 1: Find the derivative
dy/dx = -2x + 8
Step 2: Set the derivative to 0 to find the turning points
0 = -2x + 8
Step 3: Solve to find turning point
2x = 8
x = 4
Step 4: Determine nature of turning point by using second derivative
d2y/dx2 = -2
So the turning point is a maximum ☹
Step 5: Find the interval
Decreasing when x > 4
Decreasing on the right of the max, so x > 4
Finding where functions are increasing and decreasing (P32 Q2)
x = 6
Determine the x values for which f(x) = 12x - x2 is increasing.
Step 1: Find the derivative
f’(x) = 12 - 2x
Step 2: Set the derivative to 0 to find the turning points
0 = 12 - 2x
Step 3: Solve to find turning point
2x = 12
x = 6
Step 4: Determine nature of turning point by using second derivative
f’’(x) = -2
So the turning point is a maximum ☹
Step 5: Find the interval
Increasing when x < 6
Increasing on the left of the max, so x < 6