MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 2 ...

1 MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 2 advanced DIFFERENTIATION CONTENTS Function of a Function DIFFERENTIATION of a Sum DIFFERENTIATION of a Product 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. kxkx121nnakedxdy aeyx1xdxdy ln(ax)yatan(ax)adxdy tan(ax)yasin(ax)dxdy cos(ax)yacos(ax)dxdy sin(ax)yanxdxdy axy=====+== ====== 1 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 dxdududydxdy=and this is the chain rule. If we have y = f(u) and u = f (x) then we find that: (x)f (u)fdxdy = dxdu (x)f and dudy (u)f= = Put another way, we can substitute a function into another function to simplify the DIFFERENTIATION process.

2 WORKED EXAMPLE Given that y = (x2 + 3x + 1)2 find dxdyat the point x = 2 SOLUTION Substitute u = f(x) = (x2 + 3x + 1) so y = f(u) = u2 32xdxdu(x)f+== 2ududy(u)f== 3)2u(2x(x)f (u)fdxdy+= = = 2(x2 + 3x + 1)(2x + 3) At x = 2 154)34)(164(2dxdy=+++= WORKED EXAMPLE No. 2 Given that y = 4 cos(5x-2) find the equation fordxdy SOLUTION Substitute u = f(x) = (5x-2) so y = f(u) = 4 cos(u) 5dxdu(x)f== -4sin(u)dudy(u)f== )-4sin(u)(5(x)f (u)fdxdy= = = -20 sin (5x-2) SELF ASSESSMENT EXERCISE 1. Given y = (x2 - 4x + 5)4 find the equation for dxdy 4(x2 - 4x + 5)3 (2x -4) 2. Given y = sin( 2) find the equation for d dy (2 cos2 ) 3. Given y = 5cos( 3) find the equation for d dy (-15 2 sin3 ) 2 2. DIFFERENTIATION OF A SUM This is straight forwards, each term is differentiated separately so if for example y = u + v + w +.

3 Dy/dx = du/dx + dv/dx + dw/dx + .. WORKED EXAMPLE Find the gradient of the curve y = x3 + 3x2 x 1 at the point x = 2 SOLUTION dy/dx = 3x2 + 6x -1 and at point 1,2 dy/dx = 3(2)2 + 6(2) -1 = 23 SELF ASSESSMENT EXERCISE 1. Given V = 3sin( ) + 2 cos( ) evaluate dV/d at = 30o ( ) 2. Given F = + ln(2t) evaluate dy/dt at t = s ( ) 3. Given y = 5x2 + e2x evaluate dy/dx at x = 1 ( ) 3. DIFFERENTIATION OF A PRODUCT When it is difficult to multiply out an expression we can differentiate with the following rule. y = u v dxduvdxdvudxdy+= WORKED EXAMPLE Find the gradient of the curve y = (x3 + 1)(x2 + 2) at the point x = 1 SOLUTION Let u = (x3 + 1)and v = (x2 + 2) 2xdxdv 3xdxdu2== ()())(3x2x(2x)1xdxduvdxdvudxdy223+++=+= Put x = 1 13)3)(3()2)(2(dxdy=+= SELF ASSESSMENT EXERCISE 1.

4 Given y = (2x + 3)(x2 - 1) find dy/dx (6x2 + 6x -2) 2. Given F = (x2)(e2x) find dF/dx 2 e2x (x2 + x) 3. Given y = sin(2t) cos(4t) find dy/dt 2cos(2t) cos(4t) 4 sin(2t) sin(4t) 3 4. DIFFERENTIATION OF A QUOTIENTThis rule helps us differentiate a function of the form y = u/v 2vdxdvudxduvdxdy = WORKED EXAMPLE No. 4 32xx14xxy22++++= Find the gradient of the curve at the point x = 1 SOLUTION 22xdxdv 32xxv42xdxdu 14xxu22+=++=+=++= ()()()()2222232xx22x14xx4)(2x32xxvdxdvud xduvdxdy+++++ +++= = Put x = 1 ()()()()313624-36321221414)(2321dxdy2==+ ++++ +++= SELF ASSESSMENT EXERCISE No. 4 1. Given 1xxy+=find dxdy {Answer (x +1)-2} 2. Given 1x1xy +=find dxdy {Answer -2(x -1)-2} 3. Given 2x12xxy3+++=find dxdyat x = 2 {Answer -43/16} 4.

5 Given 1xey22x+=find dxdyat x = 1 {Answer } All the rules described can be combined. The final exercise requires you to do this. SELF ASSESSMENT EXERCISE No. 5 1. Given find x)3)(6cos5x(xy23++=dxdy (18x2+30)cos2(x)-2(6x3+30x+18)cos(x)sin( x) 2. Given ()324x1xey+=find dxdy ()()422x322x1x6e1x2edxdy+ += 4 5. TURNING POINTS Consider the function y = x3 5x2 +5x + 2. 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 = and but we need to use calculus to find them precisely.

6 The important thing to note is that at A and B the gradient of the graph is horizontal so the value of dy/dx must be zero. This enables us to find the value of x and y at these points. Here is how to do it. 510x3xdxdy 25x5xxy223+ =++ = At the turning points dy/dx is zero so equate to zero as follows. 510x3x0 510x3xdxdy22+ =+ = This is a quadratic equation and we must solve it to find the two values of x. Quadratic equation 2a4acbb x 0cbxax22 ==++ In this case a = 3, b = -10 and c = 5 so solving we get )(10)(2a4acbbx22 = = = = There are two possible solutions because all positive numbers have a positive and a negative square root. 40 = ===+= = = Hence the turning points occur at x = 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.

7 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 = 2x3 12x2 + 10x (Answers Min and max) 2. p = 4q3 20q2 + q +1 (Answers Min and max) 5