MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 2 ADVANCED DIFFERENTIATION

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1 MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL ADVANCED DIFFERENTIATION CONTENTS Function of a Function Differentiation of a Sum Differentiation of a Proct Differentiation of a Quotient Turning Points In this tutorial you will learn how to differentiate more complicated expressions. Below is a list of standard differentials. y ax y ae n y sin(ax) y cos(ax) y tan(ax) y ln(ax) kx n 1 anx acos(ax) asin(ax) a + atan(ax) 1 1 x x kx ake D.J.Dunn 1

2 1. FUNCTION OF A FUNCTION If a variable y depends on a second variable u which in turn depends on a third variable x, then and this is the chain rule. If we have y f(u) and u f (x) then we find that: f (u) f (u) and Put another way, we can substitute a function into another function to simplify the differentiation process. WORKED EXAMPLE No.1 Given that y (x + 3x + 1) find at the point x Substitute u f(x) (x + 3x + 1) so y f(u) u x + 3 f (u) u f (u) u(x + 3) (x + 3x + 1)(x + 3) At x ( )(4 + 3) 154 WORKED EXAMPLE No. Given that y 4 cos(5x-) find the equation for Substitute u f(x) (5x-) so y f(u) 4 cos(u) 5 f (u) -4sin(u) f (u) -4sin(u)(5) -0 sin (5x-) SELF ASSESSMENT EXERCISE No.1 1. Given y (x - 4x + 5) 4 find the equation for. Given y sin(θ ) find the equation for dθ 4(x - 4x + 5) 3 (x -4) (θ cos θ) 3. Given y 5cos(θ 3 ) find the equation for dθ (-15θ sin 3 θ) D.J.Dunn

3 . DIFFERENTIATION OF A SUM This is straight forwards, each term is differentiated separately so if for example y u + v + w +... / / + dv/ + dw/ +... WORKED EXAMPLE No. Find the gradient of the curve y x 3 + 3x x 1 at the point x / 3x + 6x -1 and at point 1, / 3() + 6() -1 3 SELF ASSESSMENT EXERCISE No. 1. Given V 3sin(θ) + cos(θ) evaluate dv/dθ at θ 30 o (1.598). Given F e -0.1t + ln(t) evaluate /dt at t 0.1 s (9.80) 3. Given y 5x + e x evaluate / at x 1 (15.545) 3. DIFFERENTIATION OF A PRODUCT When it is difficult to multiply out an expression we can differentiate with the following rule. dv y u v u + v WORKED EXAMPLE No.3 Find the gradient of the curve y (x 3 + 1)(x + ) at the point x 1 Let u (x 3 + 1)and v (x + ) dv 3x x dv 3 u + v x + 1 (x) + x + (3x Put x 1 ()() + (3)(3) 13 ( ) ( ) ) SELF ASSESSMENT EXERCISE No.3 1. Given y (x + 3)(x - 1) find / (6x + 6x -). Given F (x )(e x ) find df/ e x (x + x) 3. Given y sin(t) cos(4t) find /dt cos(t) cos(4t) 4 sin(t) sin(4t) D.J.Dunn 3

4 4. DIFFERENTIATION OF A QUOTIENT This rule helps us differentiate a function of the form y u/v v u v dv WORKED EXAMPLE No. 4 x + 4x + 1 Find the gradient of the curve y at the point x 1 x + x + 3 u x + 4x + 1 x + 4 dv v x + x + 3 x + dv v u v Put x ( + 4) ( x + x + 3 )(x + 4) ( x + 4x + 1)( x + ) ( x + x + 3) ( ) ( )( + ) ( ) 36 3 SELF ASSESSMENT EXERCISE No. 4 x 1. Given y find x + 1 {Answer (x +1) - }. x 1 Given y + x 1 {Answer -(x -1) - } 3 x + x Given y find at x x + {Answer -43/16} x e 4. Given y find at x 1 x + 1 {Answer 3.69} All the rules described can be combined. The final exercise requires you to do this. SELF ASSESSMENT EXERCISE No Given y (x + 5x + 3)(6cos x) find (6x 3 +30x+18)cos(x)sin(x) (18x +30)cos (x)-. Given 4x e y find ( x + 1) 3 e x 6e x 3 ( x + 1) ( x + 1) 4 D.J.Dunn 4

5 5. TURNING POINTS Consider the function y x 3 5x +5x +. The graph for x 0 to x 4 is shown below. At point A the graph changes from up to down and at B it changes from down to up. Such points are called turning points. Examining the graph we see the turning points occur at about x 0.6 and.7 but we need to use calculus to find them precisely. The important thing to note is that at A and B the gradient of the graph is horizontal so the value of / must be zero. This enables us to find the value of x and y at these points. Here is how to do it. 3 y x 5x + 5x + 3x 10x + 5 At the turning points / is zero so equate to zero as follows. 3x 10x x 10x + 5 This is a quadratic equation and we must solve it to find the two values of x. Quadratic equation b ± b 4ac ax + bx + c 0 x a In this case a 3, b -10 and c 5 so solving we get b ± b 4ac ( 10) ± ( 10) ± ± 40 x a There are two possible solutions because all positive numbers have a positive and a negative square root. 40 ± ± ± x.71 or Hence the turning points occur at x.71 and The graph tells us which is A and B. Note that a turning point may be the maximum or minimum value of a function, but not in this case. The extension of this work to finding maximum and minimum values is important in science and engineering. SELF ASSESSMENT EXERCISE No. 6 Find the turning points of the following functions. 1. y x 3 1x + 10x (Answers 3.58 Min and 0.47 max). p 4q 3 0q + q +1 (Answers Min and 0.05 max) D.J.Dunn 5