Differentiation

Edexcel IAL WMA12/01 /P2/Jan 2022/Q2 (Differentiation, Stationary Points & Nature)

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

The curve C has equation​​ 

y = 27x12 – x32 – 20      x > 0 

• Find​​ dydx, giving each term in simplest form.​​ 

(2)

• Hence find the coordinates of the stationary point of C.​​ 

(4)

• Find​​ d2ydx2and hence determine the nature of the stationary point of C.​​ 

(2)

SOLUTION​​ 

a- Simply diffrentiating the given equation of y.​​ 

dydx=12 x 27x12-32 x12

dydx=272 x-12 -32 x12

dydx=272x12-32x12

b-​​ To find the coordinates of a stationary point, equate​​ dsdx​​ to zero.​​ 

(Remember, to find the x-cordinate of a stationary point of the curve, simply differentiate the equation of the curve and let it be equal to zero because at stationary points, whether it is local maxima or local minima, the gradient or​​ dydx is always zero at that stationary point. Or in simple word,​​ a stationary point is a point on a curve where the curve has gradient zero. And once we have x value substitute it in the parent equation to get the y coordinate.)

​​ 

dydx=0

272x12-32 x12=0

2x12 272x12-32x12=0 x 2x12

27-3x=0

27=3x 

x=9

y=27x-x3-20 

y=279-93-20

y=34

Hence, the stationary point is​​ 9, 34 is a maximum point.​​ 

c-​​ 

(If a function​​ f(x)​​ has a stationary point when​​ x=a, then if​​ f’’a<0, the point is a local maximum)​​ 

dydx=272x-12 -32 x12

d2ydx2= -274 x-32-34x-12 

d2ydx2= -274x3-34x

x=9

d2ydx2= -27493-349

=-274 x 27 -343

=-14-14=-12

d2ydx2=-12  

∴d2ydx2<0 

Hence,​​ the​​ nature of stationary point is maximum.