Additional Mathematics Module Form 4Chapter 9- DifferentiationPage | 105 C HA P T ER 9- DIF FE REN T IA T IO N9 .1LIMITO FA F U N C TIO N E x ampl e1 : lim ( x + 2)x2= 2 + 2=4B ri e fex p l a i n at i o n:yy =x + 2420 2 x1. If x is not 2 but 1.9, 1.99 and 1.999, the value of y gets nearer and nearer to 4 but it does not exceed4.2. It is also just the same if the value of x is 2.1, 2.01, 2.001, the value of y gets nearer and nearer to 4but it does not exceed 4.3. In this situation, we can say thatx + 2approaches 4 as x approaches 2 orlim ( x + 2) =4x24. Hence, the functiony =x + 2 has limit 4 as x 2 .E x ampl e2 : 2lim x 4x2x 2First of all, factorize the numerator first.Then simplify.lim ( x + 2)( x 2)x2( x 2)lim ( x + 2)x2=2 + 2=4 E x ampl e3 : lim ( x + 1)We know that any number that divided by zero will result infinity. For example:x x1= lim ( x + 1 )= . So if we can change the equation to be like this0x

x x1= lim (1 + 1 )=0that is 1 divided by will result zero.xx=1 + 0=1For this question, at first we have to separate the terms into two fractions. Then simplify for the terms that canbe simplified.E X E R C ISE9 .1 Find:3x 2+ 2Substitute x= into1 1and becomes . We know that(a)limx x 2(b)lim (2x + 3)x 1(c)lim 2x 2x 5x 1= 0 .(d) limx4x 29x 39 .2F IR STD E R IV ATIV EO FA F U NC TIO N 9 .2 1G R AD I E N TO FTH ETAN GE N TTOA C U R V EAN DTH EF IR STD E R IV ATIV EO FA F U N C TIO N 1. Tangent to a curve is a line which is just touching the curve and not cut the curve.2. The gradient of the tangent to a curve can be determined by finding the small changes in y divided by small changes in x.B ri e fex p l a i n at i o n:o Q(1.1,2.4)o P(1,2)1. When the point Q is move nearer and neared to the point P, there will be a point which is very near topoint P but not the point P and there is a very small change in value of x and y at the point from point P.2. The point and the point P are joined in a line that is the tangent of the curve.2. The small change in y is the difference in value of y between the point and point P while the smallchange in x is the difference between the value of x of the point and point P.3. Small changes in y can be written as y that is read as delta y while small changes in x can be written as x that is read as delta x.Q(x+ x , y + y )oo P(x, y) xyx+ xThis graph show the point Q thatvery close to point P that has moved nearer to point PTangent to the curve4. From the graph above, we know thatmPQy= y . x5. But at point P, mPy=limx0 x6.limx 0 xis the gradient of tangent at point P.y dy7. To writedylimx 0 x, it is a quite long, so we can write them as .dx8. is differentiation.dx9 .2 2D IF FE R E N TIATIO NBYTH EFIR STP R IN C I P L E y = x 2o Q( x+ x , y + y )o P(x, y)y =x 21The both points lie on the same curve, sothey can be solved simultaneously.y + y =( x + x) 2y + y = x 2+ 2 x x + x 222 - 1,Left hand side and right hand side aredivided by x to make the term y . They + y y =x 2+ 2x x + x 2 x 2 y =2 x x + x 2y 2 xx + x 2= x xxright hand side is then simplified.When Q approaches P, there is no change in x or x approaches 0.y= 2x + x xFor left hand side, we already know thatlim y= lim 2 x + xlim y dy x0 x x 0 x 0 xcan be converted into . Fordxdy=2x dxright hand side, the value of x isreplacing by zero.E X E R C ISE9 .20 dyFind of the following equation using the first principle.dx(a)y =3x 2(c)y = 2x(b)(d)y =2 x 2+ 5y =5 3x(e)y =1 3x 2(f)y =3x 25x- Findingy = ax ndyby using formula -dx If the y =x 2 , then dydx=2 x .dy=a.n.x n 1dxIff ( x) = x 2 , thenf ' ( x) =2 xGiven that y = f(x),Ify =2 xd dThe first derivatedy is equivalent tof ' ( x) ,dx( y) = (2x)dx dxdyThat is iff ( x) =ax n ,dx= 2Then,Iff ( x) =2xd[ f ( x)] =d(2 x)f ' ( x) =a.n.x n 1dx dxdf ( x) =2dxdf ( x) =dxf ' ( x)f ' ( x) =2E x ampl e1 : y =x 2Compare the solution by using formula and by using the first principle. We getdy =dxd( x 2 )dxthe same answer for the same equation.dy=1.2.x 2 1dxdy=2 x dxE x ampl e2 : f ( x) =3x 2+ 6f ' ( x) =3.2.x 21+ 6.0.x 01f ' ( x) =6 xTipsWhendifferentiatetheterms without thevariableorunknown,it will result zero.E X E R C ISE9 .21 dy dy1. Find for each of the following equation. Hence, find the value ofdxat point where x =2 .dx(a)(b)y =2 x 44y = x( x 2+ 5)2 x 2+ x(c)y =x 22. Find f ' ( x) for each of the following functions. Hence, find the value off ' (3) .(a)(b)(c)f ( x) =x 2+ 3xf ( x) =(2x 1) 2f ( x) =(2x 2)( x 3)9 . 3 . 1D i ff er en ti a teex p r ess i o nwi thr es p ec t tox E x ampl e1 : We add the terms x0 because to do differentiation, it involves x if it is respectDifferentiateSolution:1 x 44 with respect to x.2to x.x0 = 1.We cannot change the expression into and( 1 x 44x 0 )dx 2equation and then differentiate it.= 4. 1 x 410.4x 0 1The solution must be started with2=2 x 3d( 1 x 44) , cannot withdx 21 x 44 .2Page | 114E x ampl e2 : Differentiate(2 x 3)(2 x + 3) with respect to x.d[(2 x 3)(2 x + 3)]dxd(4 x 29x 0 )dxFirst of all, expand the bracket. Then differentiate the expres sion.0= 2.4.x 2 10.9.x 01We add the terms x because to do=8xdifferentiation, it involves x if it is respectto x.x0 = 1.E X E R C ISE9 .22 Differentiate each of the following with respect to x. (a) 3x 25(b)x 22x(c)(3x 4)(3x + 4)9 .3F IR STD E R IV ATIV EO FC O MP O SITEF U N C TIO N E x ampl e1 : dyFind for the functiondxy =(2 x + 1) 2Solution:M eth o d1 y =(2 x + 1) 2y =4x 2+ 4x + 1dy=8 x + 4dxM eth o d2 y =(2 x + 1) 2First of all, expand the bracket. Then differentiate itLetdy duu =2 x + 1We know the value and , but wedu dxdu=2dxdyhave to find . So we have to use thedxconcept below. Page | 115y =u 2 dy=2u duIf we simplifydy =dxdy du dudxat rightdy=dy dudydx du dx hand side, it will resultchain rule.. This is calleddxdy=2u 2dx=4u=4(2x + 1)=8x + 4B ri e fE x plainat ion: From the second method above, the concept to differentiate it is:1. Consider the expression in the bracket as a term such as u but cannot x. Then differentiate it.y =u 2dy=2u duTips2. Then, differentiate the part in the bracket.u =2 x + 1We can use method 1 to differentiate if thenumber of power is small such as 2. But, if it is the power of 4, 5 and above, we can still usedu=2dxthe method 1 but it is very complicated tosolve but it is easy to differentiate by using the method 2.dy3. Multiply both of them and it will result .dx4. We can use the method 2 in all situations easily but method 1 can be used easily in certain situationsonly that is the number of power is small such as 2.E x ampl e2 : dyWe are not supposed to use method 1. It isFindLetfor the functiondxy =(3x + 4) 5quite complicated have to expand the bracketbecause it is the power of 5. So we can use the method 2.u =3x + 4du=3dxy =u 5dy=5u 4dudy= dy dudx du dxdy=5u 4 3dx4=15u 4=15(3x + 4) 4Substituteu =3x + 4 into 15uWe can also write directly like this:dy=5(3x + 4) 51 .3dx=15(3x + 4) 4E X E R C ISE9 .3 dyFind for each of the following equation.dx(a)(b)(c)y =(1 3x) 4y =(1 + 1 ) 3xy =1(3x 5) 49 .4F IR STD E R IV ATIV EO FTWOP O L Y N O MIAL S 9 .4 1FIR STD E R IV ATIV EO FTH EP R O D U C TO FTWOP OL Y N O M IAL S y =(2 x + 1)( x + 1)u vy =uv1How to get the formula?y + y =(u + u)(v + v)y + y =uv + u v + v u + u v22 - 1, ,y + y y =uv + u v + v u + u v uv y =u v + v u + u vy= uv + vu + uv x xy= uv + vu + uv x x x xlim y=lim ( uv + vu + uv )dx 0 xdx 0 x x xlim y=lim u lim v lim v lim u lim u lim vdx 0 xdx 0 dx 0 xdx 0 dx 0 xdx0 dx 0 xdy= u dv + v du + 0. dvdx dx dx dxdy= u dv + v dudx dx dxE x ampl e : dyFind for the functiondxy =(2 x + 1)( x + 1)y =(2 x + 1)( x + 1)u vu =2x + 1du=2dxv =x + 1dv =1dxThe formula isdy=u dvdx dx+ v dudxdy, just substitute each term into the formula to finddxdy=(2 x + 1)(1) + ( x + 1)(2)dx=2 x + 1 + 2 x + 2=4 x + 3E X E R C ISE9 .40 dy dy1. Find for each of the following equation.Hence, find the value odxat point where x =1.dx(a)(b)(c)y =(1 3x)( x + 2)y =( x + 2)( x 3) 2y =( x + 3) 2 ( x + 2) 32. Find f ' ( x) for each of the following functions. Hence, find the values off ' (1) and 2f ' (2)(a)(b)(c)f ( x) =(2x + 5)( x + 3)f ( x) =(2x 1)( x + 4) 5f ( x) =(2x 2) 4 ( x 3) 29 .4 .2D iffe r e nt iat ee x pr e ss ionw it hre s pe c t toxE x ampl e : Differentiate 3x 2 (2 x 5) 4 with respect to x.3x 2 (2 x 5) 4u vu =3x 2du=6x dxv =(2x 5) 4If it is given an equation, the formula used isdv =4(2x 5) 3 .2dx= 8(2x 5) 3dy=u dvdx dx+ v dudx. If it is given an expression notequation, the formula used is justu dv + v du without equal sign because it is and[3x 2 (2 x 5) 4 ]dx dxdx=3x 2 .8(2 x 5) 3+ (2 x 5) 4 .6x=24 x 2 (2 x 5) 3+ 6 x(2 x 5) 4=6x(2x 5) 3 [4 x + (2 x 5)]=6x(2x 5) 3 (6x 5)expression not an equation.Factorize2E X E R C ISE9 .41 Differentiate each of the following with respect to x. (a)(3x 5)(2 4 x) 3(b)4 x 3 (5x 6) 5(c)(8x 9) 3 ( x + 4) 49 .4 2FIR STD E R IV ATIV EO FTH EQU O TIE N TO FTWOPO L Y N O M IAL S y =2 x + 1x + 1y =uvuv1 How to get the formula?y + y = u + uv + v2 - 1, ,y + y y = u + u uv + v vy = v(u + u) u(v + v)v(v + v)v(v + v)y = uv + vu uv uvv(v + v)y = vu uvv(v + v)y= vu uv

1 x v(v + v) x u vvuy= x x x v(v + v) uy v v ulim=lim x x dx 0 xdx 0 -v(v + v)= u v v u = lim

x x dx0 -v(v + v)lim v lim u lim u lim vdx 0 dx 0 xdx 0 dx 0 xlim v lim (v + v)dx 0 dx 0v du u dvdy= dx dxdx v.vv du u dvdy = d xd xdx v 2E x ampl e : dy 2 x + 1v du u dvdyd x d xFind for the functiony =dxx + 1by using formula = .dx v 2u=2 x + 1du= 2dxv= x + 1dv=1dxdySubstitute each term into the formula to find ,dxdy= ( x + 1)(2) (2 x + 1)(1)dx ( x + 1) 2= 2( x + 1) (2 x + 1)( x + 1) 2= 2 x + 2 2 x + 1( x + 1) 2=3( x + 1) 2E X E R C ISE9 .42 dy dy1. Find for each of the following equation.Hence, find the value odxat point where x =1.dx(a)(b)(c)(3x + 2)2y =2 x 5(2 x + 1)3y =

(x + 1)2(5x 6)2y = (4 x 7)32. Find f ' (x) for each of the following functions. Hence, find the values off ' (1) and 2x + 2 x 2f ' (5)(a) f (x) =2 x 5(b)(c)(3x + 2)3y =2 x 5(3x + 2)2y =2 x 25x9 .4 .3D iffe r e nt iat ee x pr e ss ionw it hre s pe c t toxE x ampl e : Differentiate3x 2+ 2 xx + 1with respect to x.u =3x 2+ 2xdu=6 x + 2dxIf it is given an equation, the formula used isv =x + 1v duu dvdv=1dy= d x d x . If it is given an expression notdxdx v 2d" 3x 2+ 2 x equation, the formula used is just"dx x + 1 v du u dv(x + 1)(6x + 2) (3x 2+ 2x)(1)=dxdx v 2without equal sign because it is an(x + 1) 26x 2+ 8x + 2 3x 2+ 2x=( x + 1) 2expression not an equation.3x2+ 10x + 2=( x + 1)2E X E R C ISE9 .43 Differentiate each of the following with respect to x.x 2+ 3(a)(b)(c)3x 24 x 3+ xx 25x 22 x + 19 .5TAN G E N TAN DN O R M ALTOTH EC U R V E Normal to the curveTangent to the curve1. Tangent to a curve is a line which is just touching the curve and not cut the curve.2. Normal to a curve is a straight line perpendicular to the tangent at same point on the curve as shownin the figure above.Hence, since the both straight lines are perpendicular to each other,mtan gent mnormal= 1dy3. We have learned that is the gradient of tangent. So, when we are going to find the gradient ofdxnormal, at first we have to find the gradient of tangent.4. If it is given the gradient of normal, we can find the gradient of the tangent by using formula above.E x ampl e1 : Find the equation of the tangent and the equation of normal to the curve(3, 4)Solution:y =2 x 29 x 5y =2x 29x 5 at the pointdy=4 x 9dxAt point (1, 3),dy=4(3) 9dx=12 9=3Substitute the value of x into the equation to find thegradient of tangent at point (1, 3)Gradient of tangent is 3.m tan gent m normal= 13 m normal= 1m normal= 13The equation of the tangent at (3, 4) isy 3 = 3x 3y 3 =3x 9We have learned that to find the gradient, wey =3x 6use the formulay2 y1 . Use the point (x, y)x221The equation of the normal at (3, 4) isy 3 = 1that is as a general point and given point (3, 4).x 3 33 y 9 =3 xx + 3 y 12 =0Page | 131E x ampl e2 : Find the equation of normal to the curve y =3x 24 x + 2 that is parallel to the line 2 y x =5 .Solution:y =3x 24 x + 2dy=6 x 4dxWe do not know any point so we cannot find the gradient of tangent at the moment .2 y x =5y = 1 x + 5Given that the gradient of normal is equal to the gradient of this line. So we can find the2 2gradient of normal.1m =2m normalis parallel to this line, somtan gent mnormal = 1m normal=

12 .Normal to a curve is a straight linemtan gentmtan gent1 =12=2perpendicular to the tangent. We know the gradient of normal, so we can find the gradientof tangent.dyWe know that =6 x 4 andmdx6 x 4 =2x = 13tan gent=2 , henceAt the early of the solution, we gotdy=6x 4 that is the gradient of tangent. Fromdxthe gradient of normal, we got the gradient of tangent. So compare these two equationsmtan gent =2 andy =3x 24 x + 2x = 1 ,m normal=

12 at point whichx = 1 .3We are going to find the value of y for this pointthat results the gradient of tangent and normal are32" " -2 and1respectively. We substitute the value of2y =3" 1 3 4" 1 + 2 3 x into the equation of the curve that is2=1y = 3x 4 x + 2 to find the value of y. " 1Hence the point is " 3,1 ." 1The equation of the normal at " 3,1 isy 1= 1x 123Multiply the equation by 32 y 2 = x 136 y 6 =3x 13x 6 y + 5 =0E X E R C ISE9 .5 Find the equation of tangent and the equations of normal for each of the following functions at given points:(a ) y = x 36x 2+ 4 x ;x =3(b) y =( x + 2)( x 3) 2 ;x =2x 2+ 3(c) y =3x 2;x =19 .6SE CON DO R D E RD IF F E RE N TIATIO N dy1. First differentiation is .dxd 2 y2. Second order differentiation isdx 2that is we differentiate for the second times.Exa mp l e 1:d " dy d 2 y2" =2Given y = x 36 x 2+ 4x ,find d y .dx dx dxy = x 36 x 2+ 4xdy=3x 212 x + 4dxd 2 ydx 2Differentiate one more timedx 2=6 x 12=E x ampl e1 : Given f ( x) = x 3+ 4 x + 1 ,findf '' ( x) .xf ( x) =x 3+ 4 x + 1df ( x)If =dx2f ' ( x) ,dx thenf ( x)2f '' ( x)f ( x) = x 3+ 4 x + x 1dxf ' ( x) =3x 2+ 4 1x 2f '' ( x) =6 x + 2 x 3f '' ( x) = 6 x + 2x 3E X E R C ISE9 .6 d 2 yDifferentiate one more timed 2 y1. Finddx 2for each of the following equation. Hence, find the value ofdx 2at point where x =1.(a)(b)y =2x 45xy =2x( x 2+ 5x)2 x 2+ 3x(c)y =x 22. Findf '' ( x) for each of the following functions.(a)(b)(c)f ( x) =x 2+ 3xf ( x) =(3x 2) 2f ( x) =(4x 2)(x 3)39 .7C O N C EP TO FM AX IM U MAN DM IN I M U MV AL U E S yCD ABx1. Based on the graph above:(a) A and C are maximum points(b) B and D are minimum points(c) A, B, C and D are turning points2. At turning points,dy=0dx3. This is because the line of the tangent at the turning points is a horizontal line that is parallel to x-axiswhich is the gradient is 0.dyis the gradient of tangents so at at turning pointsdxdy=0 .dx9 . 6 . 1S tep stod eter mi n eatu r n i n gp o i n t i smax i mu mo rmi n i mu m dy1- Finddx2- Determine the turning points, find the value of x then find the value of y3- Determine whether the turning point (x, y) is maximum or minimum.(a) minimum point:x x1dydxnegative 0 positiveSketch position oftangentShape of the graph(b) maximum point:x x1dydxpositive 0 negativeSketch position oftangentShape of the graphE x ampl e1 : Find the turning point for the function y = x 33x 29x + 5 . Hence, state each point is minimum or maximum.Solution: M e thod1 y = x 33x 29x + 5dy=3x 26 x 9dxx 0 ,d 2 y< 0dx 2then it is a minimum value.(-1, 10) is a maximum pointAt point (3, -22),d 2 ydx 2= 6(3) 6=12d 2 y> 0dx 2(3, -22) is a minimum point.E X E R C IS E9 .7 1. Find the turning point for each of the following function. Hence, state each point is minimum or maximum.(a ) y = x + 1x216(b)f ( x) = x +x 22. Find the turning point at the following curve(a ) y =8x x 2296(b ) f ( x) =2 x +x 29 .8R ATEO FC H A N G E 1. The first derivative for a function denotes the change in the quantity y with respect to the change in the quantity x." d2. Rate of change is differentiation that is respect to time " 4. The formula for rate of change isdA = dA dB dt dt dB dtwhere A and B are variables that can be changes depends on the case and situation.E x ampl e1 : The sides of a cube increases at the rate of 1.4cms-1. Find the rate of change of the volume when the sides measure 5 cm.Solution:In this case, A is the volume (V) and B is the side(s).Given thatds = 1.4cms 1 ,s = 5cm and V= s 3 . We are going to find dV.dt dtV=s 3dV=3s 2dsV=s 3 is the formula for the volume ofcube. Differentiate it that is respect to s.The formula isdV= dV dsdt ds dtdV2Substitute =3sinto the formula.dsdV=(3s 2 ) dsdt dtWhends=1.4cms 1 ,s=5 ,dtdV=3(5) 2 1.4dtSubstitute the value of r that is 5 into 3s2ds=75 1.4=105cm 3 s 1and the value ofdtinto the formula.E x ampl e2 : The volume of a sphere decreases at the rate of4cm 3 s 1 . Find the rate of change of the radius of sphere when the radius is 3cm.Solution:In this case, A is the volume (V) and B is the radius (r).dVGiven thatdtV= 4 r 3=4cm 3 s 1 ,r=3cm and V=4 r 3 . We are going to finddr .3 dt43dV=4 r 2drV= r 3 is the formula for the volume of3sphere. Differentiate it that is respect to r.The formula isInfodV= dV drdt dr dtThe value ofdVis negative because the ratedVSubstitutedr=4 r 2 into the formula,dtis decreasing. If it is increasing, it will be positive.dV=4 r 2 drdt dtdVWhendt=4cm 3 s 1 ,r =3cm ,4 =4(3) 2drdtSubstitute the value of r that is 3 intodVdr =4cms 14r 2and the value ofdtinto thedt36formula.= 19cms 1E X E R C ISE9 .8 1.The area of a circular water ripple expands at the rate of6cm 2 s 1 when the radius is 5 cm. Hence, find the rate of change of the radius if the ripple.42. Given that y =5x +x when x= 4.. If y increases at a constant rate of 3 unit per second, find the rate of change ofx9 .9SMAL LC H A N G EAN DAPP R O XIM ATIO N 1. We have learned that delta x( x ) is the small change in x and delta y( y ) is the small change in y.2. We know thatlim y =dy . x 0 x dx2. The formula for small change and approximation isy dy x dxWhere x and y are variables that can be changes depends on the case and situation.E x ampl e1 : Given the radius of a circle increases from 4cm to 4.01.cm. Find the approximate change in its area.Solution:In this case, A is the area (A) and B is the radius (r).Given that r =0.01cm ,r =4cm andA = r 2 . We are going to find A .A = r 2dA =2 r drA = r 2 is the formula for the area of circle. Differentiate it that is respect to r.The formula isA dA r drSubstituteA 2r rdA =2 r into the formula,drMove r to other side. A 2 r rWhen r =0.01cm ,r =4cm , A 2(4) (0.01)8(0.01)0.08 cm 2Substitute the value of r into and the value of rinto the formula.E x ampl e2 : Given y =3x 2+ 4 . Find the approximate change in y when x increases from 2 to 2.03.Solution:In this case, A is y and B is x.Given that x =0.03 ,x =2 andy =3x 2+ 4 . We are going to find y .y =3x 2+ 4dy=6 x dxThe formula isy dydyFinddx x dxSubstitutey 6 x x y 6 x. xdy=6 x into the formula,dxMove x to other side.When x =0.03 ,x =2 , y 6(2) .0.030.36E X E R C ISE9 .9 Substitute the value of x into and the value of xinto the formula.1. Given y =2 x 37 x 2+ 5 , find the value ofwhen x decreases from 2 to 1.97.dyat the point (3, -4). Hence, find the small change in y,dx2.Find the small change in the area of a circle if its radius increases from 5cm to 5.02cm.C H AP TE RR E V IE WE X E R C ISE 1. Evaluate the following limits:4 x + 6(a)limx

5 x 32(b)lim ( x4 )x2(c)limx 21 + xx 3 1 x2. Given y=3x 24 x + 1 . Find dy .dx3. Differentiatex 35x 2with respect to x.4. Given f ( x) =( x + 2)( x 21) . Findf ' ' (3)5. Given y =(3x + 2) 5 . Find the gradient of the curve at the point where x= -1.6. Given the gradient of the normal to the curvey =kx 23x + 4 at x= -2 is 113. Find the value of k.7. Find the equation of the tangent and the equation of normal to the curve8. GivenP = xy andx + y =30 , find the maximum value of P.9.P QX cmy =3 x 2 at point (2, -1)R Sy cmThe diagram above shows a circle inside rectangle PQRS such that the circle is constantly touching the two sides of the rectangle. Given the perimeter of PQRS is 60 cm."+ (a) Show that the area of shaded regionA =30x " 44x 2(b) Using =3.142 , find the length and width of the rectangle that make the area of the shaded region a maximum.10. Given y =2x 37 x 2+ 5. Find the rate of change in y, at the instant when x= 3 and the rate ofchange ion x is 5 units per second.11. Given p = (t 1) 3+ 9 t 2 . Finddp dpand hence find the values of t where =9 .2 dt dt Page | 130 3 q 2 19212. Given that graph of functionand q are constants, find(a) the values of p and qf ( x) =px+ has a gradient functionx 2f ( x) =6x where px 3(b) the x-coordinate of the turning point of the graph of the function.13. The straight line4 y + x =k is the normal to the curve(a) the coordinates of point E and the value of k(b) the equation of tangent at point Ey =(2x 3) 25 at point E. Find14. Differentiate the following expressions with respect to x.(a)(1 + 5x 2 ) 33x 4(b)x 4+ 215. Given f ( x) =5x 31 , findxf ' ' ( x) .16. Given thatdy(a) Finddpy =2x 23x and p =5 x .when x= 2,(b) Find the small change in x when p increases from 4 to 4.05.Page | 131