Leaving Cert Differentiation
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1 Leaving Cert Differentiation Types of Differentiation 1. From First Principles 2. Using the Rules From First Principles You will be told when to use this, the question will say differentiate with respect to x from first principles. What to do step by step: 1. f(x) Write the equation that you re given 2. f(x + h) Add h to all x s in the starting equation 3. f(x + h) f(x) Take the starting equation from step 1 from the equation from step h Divide the equation from the end of step 3 by h 5. lim Change any remaining h s to zero h 0 h Differentiate 2x 2 5x with respect to x from first principles 1. f(x) = 2x 2 5x 2. f(x + h) = 2(x + h) 2 5(x + h) f(x + h) = 2[x 2 + 2xh + h 2 ] 5(x h) f(x + h) = 2x 2 + 4xh + 2h 2 5x + 5h 3. f(x + h) f(x) = 2x 2 + 4xh + 2h 2 5x + 5h (2x 2 5x) 4. f(x + h) f(x) = 2x 2 + 4xh + 2h 2 5x + 5h 2x 2 + 5x Cancel what you can f(x + h) f(x) = 4xh + 2h 2 5h 5. h = 4xh+2h2 5h h = 4x + h 5 Rules 6. lim = 4x = 4x + 5 h 0 h Rule 1: General Rule y = ax n = nax n 1 Multiply by the power and reduce the power by 1, if there is a number in front of x multiply the power by this number and reduce the power by 1 Note: The derivative of a constant number is 0 Examples: 1. y = x 3 multiply by the power 3x 2 and reduce the power by 1-3x 3 1 = 3x 2 2. y = 3x 2 = 2 3x2 1 = 6x 1 = 6x
2 3. x 3 3x + 4 = 3x2 2 3x + 0 = 3x 2 6x Rule 2: Product Rule Use this when you have to get the derivative of two expressions multiplies together i.e. two brackets multiplied together. First multiplied by the derivative of the second plus the second multiplied by the derivative of the first (First Derivative Second) + (Second Derivative First) (x 2 + 3x)(x 3 3x + 4) Break up into 1 st x 3 + 3x and 2 nd x 3 3x + 4 Get the derivative of each: Derivative of First - = 2x + 3 Derivative of Second - = 3x2 3 Apply Rule: 1st derivative 2 nd + 2 nd derivative of 1 st Step 4: Multiply out and simplify Rule 3: Quotient or Fraction Rule Use the rule: Break up expression into Top and Bottom (x 3 + 3x)(3x 2 3) + (x 3 3x + 4)(2x + 3) x 2 (3x 2 3) + 3x(3x 2 3) + 2x(x 3 3x + 4) + 3(x 3 3x + 4) 3x 4 3x 2 + 9x 3 9x + 2x 4 6x 2 + 8x + 3x 3 9x x 4 + 2x 4 + 9x 3 + 3x 3 3x 2 6x 2 9x + 8x 9x x x 3 9x 2 10x + 12 Bottom Derivative of the top Top Derivativeof the Bottom (Bottom) 2 3x + 2 x + 1
3 Top: 3x + 2 and Bottom: x + 1 Get the Derivative of both the Top and the Bottom Top Derivative: = 3 and Bottom Derivative: = 1 Apply Rule: ( x + 1)(3) (3x + 2)( 1) (x + 1) 2 3x x (x + 1) 2 = (x + 1) 2 You can leave the bottom as it is you don t need to multiply it out Rule 4: Chain Rule This is when something in a bracket is raised to a power i.e. (x 2 3x + 4) 4 Treat what s in the bracket as one, then multiply the bracket by the power and reduce the power by 1 Get the Derivative of the bracket Put it all together Evaluating the Derivative (x 2 3x + 4) 4 = 4(x 2 3x + 4) 3 = 2x 3 4(x 2 3x + 4) 3 (2x 3) When you have to evaluate a derivative get the derivative first and then sub in the value you re given for x If y = 2x 3 + 4x 2 + 3x 5, Find the value of when x = 2 Get the derivative first
4 y = 2x 3 + 4x 2 + 3x 5 = 6x2 + 8x + 3 Step 2 Evaluate x = 2 (Sub in your x value) 6(2) 2 + 8(2) = 43 Tangents You can be asked to find the equation of the tangent to the curve. (the derivative) is the slope of a tangent to any point on the curve. To find the slope and equation of a tangent to a curve at a given point on the curve there are 3 steps: Find the Derivative Evaluate the derivative at the point. Let all the x s in the derivative equal the x point so if you have to evaluate the derivative 2 2x and the point is (2,3) you get 2 2(2) = 2 4 = 2, 2 is the slope. Find the equation of the tangent using the point you ve been given and the slope you found in step 2 and the formula (y y 1 ) = m(x x 1 ). Sometimes you ll be given the f(x) and the slope of the tangents and asked to find the co ordinates of the corresponding points on the curve Find the Derivative Let the derivative equal to the given slope and solve for x When you have the x value go back to the original function and sub in your x value from step 2 to get the y value NB The Tan of an angle is equal to the slope If a line is at an angle of 45 then the slope of that line is the Tan 45 = 1 Maximum and Minimum Points You may be asked to find the Max and Min values:
5 Get the Derivative Let the derivative equal to 0 (zero) to get the x value Sub in x values into your starting equation to get the two y values Step 4: Compare the y values of each and the point with the largest y value is the maximum point, it s the same for the minimum value when you compare the y values the point with the smallest y value is the minimum point. The minimum and maximum points are also the turning points (where the graph turns). Increasing and Decreasing - If the slope of a tangent is positive, is greater than 1 - If the slope of a tangent is negative, is less than 1 Rates of Change dh dt dr dv is the rate of change of y with respect to x is the rate of change of h with respect to t is the rate of change of R with respect to V If you are given a function involving s, p and t, and asked to find the velocity and then the acceleration. s = t 3 9t t 3 The Velocity is the 1 st derivative i.e. Velocity = ds dt ds dt = 3t2 18t + 15 The acceleration is the 2 nd derivative i.e. acceleration = d2 s dt 2 d 2 s dt 2 = 6t 18 If you re asked to find the value of t when the velocity is a certain amount. Let the velocity equal the value you re given So if the velocity is 0 t = 3t 2 18t + 15 = 0 Solve for t t 2 6t + 5 (t 1)(t 5) = 0 t = 1 or t = 5 If you re asked to get the acceleration after a certain time. Sub in the time you re given for t e.g. you ve to find the acceleration after a certain time say after 4.5 seconds Take the acceleration (2 nd Derivative) 6t 18 and sub in 4.5 for t 6(4.5) 18 = = 9 and then add the units m/s If you re asked to find the time it is at a certain acceleration i.e. 6m/s let the acceleration (2 nd Derivative) equal what acceleration you re given and solve for t
6 6t 18 = 6; 6t = ; 6t = 24; t = 4 so the answer is 4seconds If you re asked to get the velocity at a certain time sub in the time you re given into the velocity. 3t 2 18t + 15 and t = 4 3(4) 2 18(4) + 15 = = 9m/s You could also be given a function h, here h is the height, t is the time Speed is the 1 st Derivative. There is no 2 nd Derivative when the question is about height.
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