Differentiation

Basic rules of differentiation

Differentiation tells us about the gradient of tangents, the derivative and the rate of change.  To differentiate: Multiply the number at the start by the power and reduce the power by 1.  If you are asked to differentiate f(x) then your answer will be in the notation f ' (x) If you are asked to differentiate y then your answer will be in the notation dy/dx    Basic examples:  f(x) = x^2                          f(x) = x^3                  y = 5x^3                      y = 9m               f(x) = x^-5 f ' (x) = 2x                         f ' (x) = 3x^2            dy/dx = 15x^2             dy/dx = 9           f ' (x) = -5x^-6   Basic examples (With fractions as the power):  f(x) = x^1/2                                                                             f(x) = 1/2 x^1/2 + 1/6 x ^-1/7 -3 f ' (x) = 1/2 x^-1/2 (1/2 x ^-1/2   not 1/2x^-1/2 )               f ' (x) = 1/4 x^-1/2 -1/42 x^ -8/7 

Differentiating products and quotients

y = (x + 3)(x + 5)               y = (square root of x + 1/x)(square root of x + 1/x)                  y = x^3 + 5x^2 - 6 all over x  y =  x^2 + 8x + 15            y = x + 2rootx/x + 1/x^2                                                               y = x^3 /x + 5x^2 /x -6/x dy/dx =2x + 8                  y = x + 2x^-1/2+ x^-2                                                                     y = x^2 + 5x - 6x-1                                          dy/dx = 1 - x^-3/2 - 2x^-3                                                                dy/dx = 2x + 5 + 6x^-2                                                     = 1 - 1/rootx^3 - 2/x^3                                                                  = 2x + 5 + 6/x^2   Note: I have done a stage extra so the powers are positive. I also understand it may be hard to understand as the computer does not properly show powers etc, so I have done the working by hand and put the answers at the end of the power point/ note. 

Applications of Derivatives

Differentiation can be used to solve real-life problems such as rate of change or gradient Example: Given f(x) = 5x^3,find the value of f ' (2) f(x) = 5x^3 f  ' (x) = 15x^2 f ' (2) = 15(2)^2 = 60  Steps for rate of change ^:  Differentiate as normal Then sub in x = whatever, in this case 2 Answer  Rate of change is usually subbing in. 

Equations of tangents

Since a tangent is a straight line, its equation may be given by y-b=m(x-a) To find the equation of a tangent at any point on a curve, we need to determine:  The coordinates (a,b) of the point  The gradient,m, at that point  Example:  Find the equation of the tangent y=rootxcubed at x = 9  y = x^3/2 (Another way of writing rootxcubed) f(x) = x^3/2 f ' (x) =3/2 x^1/2 = 3/2 root x when x = 9, y= root9^3 root9^3 = 27  point is (9, 27)  Now sub x into f ' (x) to work out the gradient f ' (x) = 3/2 x root9 = 9/2 Therefore. gradient = 9/2 Now sub in your points and gradient into; y-b=m(x-a)  y - b = m(x - a) y - 27 = 9/2 (x - 9) 2y - 54 = 9x - 81 2y = 9x - 27  ( You can leave the answer like this but i strongly recommend making it y=.... not 2y = .....) y = 9/2 x -27/2 

Hand written solutions for examples (Slide/note 1)

Hand written solutions for examples (Slide/note 2)

Hand written solutions for examples (Slide/note 3)

Hand solutions for examples(Slide/note 4)