Differentiation

Find the derivative f’(x) if :

(a) f(x) = 2x-3 (b) f(x) = 0.4 x ½

(c) f(x) = (d) f(x) = + 2

(e) f(x) = (f) f(x) =

The function is f(x) = Find an expression for f’(x)

The function is f(x) = f’(x) ?

f(x) = x3 + 2x2 – 3x + 5f’(x) = 3x2 + 4x - 3

f’(1) = 3 x 12 + 4 x 1 – 3 = 3 + 4 – 3 = 4

If f(x) = x3 + 2x2 – 3x + 5 find the value of the gradient when x = 1

Find the gradient of the curve f(x) = x3 – x2 +1 at the point where x = 2.

f(x) = x3 – 8x2 + 5x + 14

What is the gradient of the curve f(x) = ( 2x – 1)2 at the point ( -2, 25 ).

Find the equation of the tangent to the curve y = x3 + 3x2 + x - 5 at the point (-1, -4 ).

y = x3 + 3x2 + x - 5 = 3x2 + 6x + 1 find expression for gradient

At (-1, - 4 ) , x = -1

= 3 (-1)2 + 6 (-1) + 1 finding gradient at -1 = -2 gradient at the point

y – y1 = m ( x – x1) equation of a straight liney - - 4 = -2 (x - -1) substitute point and slopey + 4 = -2x – 2

2x + y + 6 = 0 or y = -2x – 6 the equation of the tangent

Find the equations of the tangents to the curve f(x) = x2 + 3x + 2 at the points where the curve cuts the x-axis.

The curve cuts the x-axis when f(x) = 0 x2 + 3x + 2 = 0( x + 2) ( x +1) = 0Either x = -2 or x = -1 the curve cuts through at both points

f(x) = x2 + 3x + 2 f’(x) = 2x + 3 differentiate to find the gradientAt x = -2, f’(-2) = -1 gradient of tangent at x=-2 is -1At x = -1, f’(-1) = 1 gradient of tangent at x=-1 is 1

Equations of the tangents using y – y1 = m ( x – x1) are( -2 , 0 ) ( -1, 0 )y - 0 = -1 (x - -2) y - 0 = 1 ( x – -1 )y = - x – 2 y = x + 1 Are the equations of the tangents where the curve cuts the x-axis.

1. Find the equation of the tangent to the curve y = x2 – 5x + 4 at the point (4, 0)

2. Find the equation of the tangent to the curve y = x3 – 3x2 + 1 at the point (1, -1).

3. What is the equation of the tangent to the curve y = -x3 + 2x2 + x – 2 at the point where the curve cuts the y-axis?

4. Find the equations of the tangents to the curve y = x2 + x – 12 at the points where the curve cuts the x-axis.

1 2 3 4 5 6 7 1 2 3 4 5 6

y

x

1

1

2

2

3

3

4

4

5

5

6

6

7

7

1

1

2

2

3

3

4

4

5

5

6

6

2503 xxxf .)( 10x15x3x)x(f 23

. Find a point, given the gradient .

To find the point(s) given a gradient we:Differentiate, and solve for x

Find the coordinates where the function has the gradient of 2

(i) derivative is

(ii) find where

Substitute this ‘x’ value back to find the ‘y’ value to get coordinates (x,y)

10x15x3x)x(f 23

Find the coordinates where the function

has the gradient of 9

(i) Derivative =

(ii) Find x where f’(x) = 9

. Find a point, given the gradient .

Second DerivativeDifferentiate twice. Use the second derivative to identify features of graphs: where the graph is increasing, decreasing, points of inflection and stationary points.

Use calculus to find local maximum, local minimum, and points of inflection.Finding the nature of Turning pointsNOTE: determining the nature of turning points you can use any method: 1. The second derivative, 2. 2. By inspection of the graph of the function or 3. 3. By testing the value of the function on each side of

the turning point

𝑦= 𝑥3

3+2 𝑥2+3 𝑥

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