Differentiation Revision Worksheets

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fI\ au t"l 3 1

pOwqs

1

ii ?,1; -- x6

12

iii {7 = (x'12 - Ysalv

rational

gtr Change to index form:

L

Xuna

-^.aLI

--n"

1-z-^/{-L

trl x'

f|DifferentiatewithresPecttox:

i *oZ) -

2*1 ii y = +!.

1,'=x4

du 1-1/ =-X +

dx4

Review of Index Laws:

fl* X On = Affi+n A* : a' = A*-' (a*)' = A*

a*bn-(ab)* ao=\ a-"=7

Find the derivatives of the following functions'

1

ttz"2 x;

1

,-r'6 7a

2-510x

simplity the followin$, and

xLarLr18 4lli

22 *{i

zG *{ixx3

34

1

=,rt x3

1

YE Zxz3

ll 4x- +

then find their

15 x3 *xa

T9

23

27

3I

35

,*

,r8t2 +*-'t

derivatives:

2;xr

L

i''1 Yz .

2572

1

g x-s

Use the index laws to

Ig x3 xxs

t7 gJ-x

T6

32

36

40

3

3x+

xa

7*Jix'Lli

x

TG1

ll x

1,

-3xs

J;-

2t x'Jix

25 4x

xz2ew

r33 lx

x

2

30 x3

20

24

28

{.

x'Jix2

GJi4

_7X3

g{i

5

1

gg xz +x-s3

G7 x+ {*

4r F,li-]-x

38 Ji + 4xz

42 {;-+{;

49 Find the gradient of the tangent to the curve y = Ji+ 3 at the point (4' 5)'

44 Find the gradient of the tangent to the curve A =1li - 4 atthe point where x = L'



dilferentiation - product rale

derivative of a product is given W fr=u#*ofr*fr"ru U=uxv.

may choose to remember this rule as:

first (factor) by the derivative of the second plus the second b

Differentiate y -(x+3X5x +2)

#= @ +3) x *rr. + z)+ (5x + 2)" *(r + 3)

- (x+ 3)5 + (5r +2)I

-5x+15+5x+2- 10x+L7

Note: We could have expanded first and then differentiated.

Differentiate f (x) - (3xz + 4x+ 3)(x' +7)

f '(x) - (3x2 + 4x+ 3) xZx + (x' +7)(6x + 4)

=6x3 +8xz +6x+6x3 +4x2 +42x+28

-I2x3 +\2xz + 48x+28 or

:4(3x3 +3x2 +72x+7)

,Sw'L g *as #

e derivative of the first'.

+ vurt,5: [LV-

I

The

You

'the

gtr

U,pe the product rule to differentiate:

.7,",,/ -(x'+2)(2x+4)-i

r 3\ff

= (xt + $(2x- 3)

y-(xn+x)(2x+9)

,7 y= (x' - 4)(x' - 2x)

,"{ y-(xt-3x2)(xn+6x)

\, Differentiate:

v*{, (2x+ 5)(rt + 4)

,13 (x'+ 6)(x' + 4)

il5 (x' +3x + 1Xx + 6)

Fil{ f '@) for the following:

/.,

/ Y -(x' +3)(r-5)

*' y -(x' -3rXx+6)it/

"eY - (xn +3x2X3r -9)

vX y-(2xt+4x)(xn-x)

,-L5. y -(xt - 4x)(2xz - x)

'/;12' (7 x' + 3x)( 2x - 3)

-/fi (x'+x)(xt -x)

4ar\

V

, 16 (xt - 3x)(2x2 + 3x)

\

./

,Vt f (x) - (6x3 + sxX x' - 1)

uLO f (x) - (4x - 3x2)(2x'+ 8x)

2I f (x)-(2x3 +3x+1)(x -2)

20 f (x) = (1 \xz - 2)(3x3 + 4x')

22 f (x)-(3x3 -2x2 +1)(ax-3)

23' lf f (x)=(x2 +2x)(x- 1), find the derivative of f (x) and hence the value of f '(3).

24 Given f (x)=(3x2 +4x)(x+3),frnd' f '(x)and hence the value of f'(1).

25 lt f(x)=(x'-3x)(x-xz),hndthevalue of f'(0)+ f'(L).

26 lf f(x)

=(x

-3x2)(2

-xa), hndthe value of

f'(-L) +

f'(I).27 II f (x\=(xz +2x-l)(x- 1), find the value of f '(1)- f '(-I).

2a Findthegradientof thetangenttothecurve y=(3x+4)(x2-2x)atthepointwhere r=1.

29 Find the gradient of the normal to the curve U =(x' +3x)(2x-3) at the point where x=2.

30 Given y =(xt -3x)(x2 -2), findthe equation of the tangent at the point whete x=2-

@ Kevin Dunne and Blake Education 2006 Senior Maths - Calculus



1 a 1I-

If u-Lth"n the derivative is +-W'ta

Lt'rr Y %U^dx u2

otient ruledu da

(2x -1) xLx - (x2 - 3)x2(2x -1)'

4x2 -2x-2x2 +6

liation

'lou may choose to remember the rule as'bottom times the derivativeof the

topminus the top times

the derivative of the bottom, divided by the bottom squared'.

(2x -L)'

2xz -2x+6

08

/t /{oc>L*u-& ? t-"r'-a"d-*Q-'

I -QS RTfi

&L/ - $*'"''

(2x -I)'

-, x2 -3.t'or y - zrL,

dy_dx

o

CI

Find the derivatives of:

7y-+c)cz - x

x2 +I10 y=-: x-2

/Differbntiate:

at'xz +1.

13-x"

2xz7,6

"x, _1.,/

.,///

vyg Find the

24

25

-4x2 --=./ r 3x+2

,/' Xz.A a-/ - -/ 4x-5

6 lI=ra_-'- ,/x2 _ 2

3-x9 Y= 4x-s/

az' tr -3-x27- '/ 4-2x

Axz15,

6x+5

-r ,/L

16.xr +1,

,,,,'8

Y-x2+1

t'3x +1

4t

-Y - 2x+!,' x2 -3aI u-lr- v'

x +'l'

/,'/-- X

x3

x2 +1,

x3

T4

a72x2 +L

gradient of the tangent to the curve , =# at the point where x = 1.r -pt:.

// , *-L

#( Find,the gradient of the tangent to the curve , = 4 at the point where r = 3."x-2'

2t Determine the angle of inclination (to the nearest degree) to the positive x-ais, of the tangent to

x+It

-''/ x +3

"-'''

at the point where x ='/-,. . ,

22 Determine the angle of inclination (to the nearest degree) to the positive x-axis of the tangent to

..."-t - ,/

12 -tl *. ? ".r +- ,3 *,..''.' ',

y=-:-atthepointwherex=1. I i' ' "

x-L

29 Find the equation of the tangent to the curve , =*at the point (-3,3)."x+1^!

Find the equation of the tangent to y ==at the point where r = t. 'ia'"

3- 4x

Findtheequationofthenormaltothecurve r={-atx=1.* f t ?- #'tt"' L '-

Given f (x)='$determine the equation of the tangent and normal to the curve at thex+L

vrY.

- 2x+I

,/ 4'- -t-4 Y= gx+4

26



1'

ditterentiation - chain rule

The Chain Rule for differentiatirg a function of a function is:

If y = f (u) and u- f (x) then + -+"+ dx dudx

Use the chain rule to find

+ff y

-(xu

-4)n

dxv

Let n = x3 -4 thu" + -- 3x2 and A = ua thus + = 4Lt3

dx 24 du

U-U*d'dx du dx

fu- = 4tt3 x3x2 but u - xs - 4dx

ry - A(xs - 4)' xzxzdx\

- 12xz1xt - 4)t

g

tr

E

Students are encouraged to perform this differentiation using lhe shorter method below:

Differentiate J 4x 1

**n.-3); =;@x4t;x4

=2(4x4)-+2

Find f'(x) given f (x) - (3xz - 4x)u

f '(x) = 5(312 - 4x)n x (6x - 4)

= 5(6x- 4X3 xz - 4x)n

4x-3

Differentiate the following:

v'! Gx+z)u

/4 (7x-3)u

y 7 (2- 3x')n

,rl0 (3x- 1)t

fr J4x+2

to (7x-3)'

rs 4J6,;7

(5x + 3)n

(x2 + 4)3

(3xn + 6)u

L

g4r,

3

6

I

T2

(7 - 6x)n

(4x'-7)'

(7 x'- 3)-'

7

Zxz + l-

&-zr'z

{m3

ffi;22 .Given f(x)=(3-x')t,findthe value of f'(2).

;Z{" Given f(x) = (4-3x2)a,find the value of f'(3).2+ Given f(x)=J*'-l',findthevalue of f'(4).

]6 .petermine the gradient of the tangent to the curve y = (3x - 2)3 at the point where x = 1.

"

2{ finathe gradient of the tangent and the normal to the curve y =(xz +l)a at the point (L,L6).

2i" Ftndthe equation of the tangent and the normal to the curve U =(2- xz)s at the point (1,L).

2

I

5

8

17,

a7 (6 - x)'

20 sJ64

15

18

2t

@ Kevin Dunne and Blake Education 2006 t3 Senior Maths - Calculus



lls te ien 2

tr y=Ji a f@)-L {._l

f(x)= x3

'f '(x) ! *-i3

-Lg1'l xn

E y=(x'+3)(x'+1)

y' = (x' +3)3x' +(xt +1)2x=3xa +9xz +2xa +2x

= Sxa +9xz +2x

I!=xzdy

-'J' -.-+W

dx21"

ffitrf (x) - (2x - x')'

f '(x) = 3(2x - x2)'(2-2x)

3(2-2x)(2x- x')'

x x3(2x + 1)t x2- (2x + 1)3 x 1

xz

(2x +1)'(6x -2x- 1)'x2

(2x+ 1)'(4x- 1)

x2

(2x+ t)t4t

-

\

9t-rx

Differentiate:2

Ivi

(3 + 2x)'

7 (x' -3)(x' + 4)

11O-.^11x'-5

4xI3

xz +3

Findfu-

for the following functions:dx

(x+3)(x'-2)

1.

x+g3x

-x-3

(3x' -I)(zxt +2)

(3 - 4x')'

2x

x+L

(3 - 4x)n

x'Ji

L

Ag y=(x'+3)u(3-x')

22 y-(xt+3x+4)n

25 u= x- r (x -r)'

18 l,t -x2

r 2-3xz

2r y_h

24 y-(x'+3x)(x3-L)'

6 7Ji

12

153x-x2

rG y -*{i 17 y -(x'-x)(xt +1)

220 Y= f +gx

223 Y- f -4,x

2x26 y _(tr+1f

27 Find the gradient of the tangent and normal to the ct3

nve y = f* uf the point where r = 9.

28 Find the gradient of the tangent and normal to the curve , =fiat the point[t,i)

29 Determine the equation of the tangent to the curve y =(x'+1)(2x- 3) at the point x = 1.

30 Given f (x)=(x'-4)', determine f'(I)- f'(-1).



atOcaiil (relativ{ ma*ima and minima and horizontal points

otl iiltexion asing the lirst derivative - conlinaed

the nature of each of the six different stationary points at x=3.

x 2 3 4

f '(x) >0 0 >0

x 2.9 3 3.1_

f '(x) <0 0 <0

Given the following information, state the nature of the stationary point at the point where )c = a..

7 f'(a)-o 8 f '(a)- o e f'(a)-o

f'(o+1)>0

f'(o-1)<o

10 f '(a) - 0

tt Show that y = xz + 4x+ L has a minimum turning point at (-2,-3).

t2 Show that y =-x2 +6x,- L has a maimum turning point at, (3,8).

13 Show thal y = x3 + 4 has a horizontal point of inflexion at the point (0,4).

14 Show that y = x4 + 4 has a minimum turning point at (0,4).

15 Determine the coordinatds of the stationary point of the curve ! = -x4 -6.16 Determine the coordinates of any stationary points of y = xz +8x+5 and state their nature.

t7 Find the coordinates and the nature of any turning points fot y - 4x2 + 4x+3.

18 Determine the coordinates of the stationary points of. y =2x3 -6x and state their nature.

t9 Show that the curve U =2x3 +15x2 +24x+12has 2 stationary points. Find the coordinates ofthe points and determine their nature.

20 The curve A = x3 +3x2 +3x+'J. has one stationary point. Determine its nature.

2l Show that y =(2x-3)3 has a horizontal point of inflexion. Determine its coordinates.

Given the following tables, state

a

)c 2.25 3 3.L

f '(x) <0 0 >0

22 Determine the coordinates of the stationary points on thedetermine their nature.

Determine the nature of the stationary points on each23 A:-x2+4x+524 U:x2+2x-625 y - x3 -3x26 y =2x3 +9x2 + 48

16Acurve y - ,x3

-xz -3x+4 and

4

6

f'(,.;)<0

f'(, i) <0

f'(,-i) >0

f'(, i) <0

t'(o-i) >o

r'('-!n)>o

)c

2!23 31

2

f '(x) >0 0 <0

x, 2 3 4

f '(x) >0 0 <0

x 2!2

3 3!2

f'(x) <0 0 <0

Senior Maflrc - Cnln',ln,o

of these curves.



&,'.-'"a' j {. , ''i '

11 r. "r "-r/ ../

;ima and minima problems

,,talculus may be used to rlhr" maxima and minima problems.

EE! A-preschoo-I owner wants to make a secure play area for'his preschoolers. He has 60 metresof fencing. If he uses an existing side fence,-what is the *"*i-.r- area he

"ur, "rr"ior"zLet I be the length of

Then:

From the diagram:

(1)

(2)

Maximum area occurs whenb = L5.

the play area and b the breadth. Existing fence

A-Ixb

l+2b:60

From (2): I = 60 -Zb

Substitute into (1) 2 A= (60b -2b)xb

A- 60b -2b2 (3)

Now we aflPly the Process for finding maxima/minima.

#=60 -4b

LetL Q

db

d2Arw - -{ which is < 0 Area is a maximrrm.

d

i.e.50-4b=Ab=15

To find the area, we substitute into equation (3):

A-60x15-2x1,52

= 450 Maximum area enclosed is 450 mZ

A-piece of rope is used toa 24:-metre piece of rope.

*'2 A farmer wishes to form a,temporary_paddock using 100 metre of wire. If he uses an existing{ence as one of the sides, determineilie maxim.r- u-r"" of the paddock.

Y Two numbers add up to 32. Find the numbers, if the sum of their squares is a minimum.

,/ The total daily profit of a factory,which produces television sets, is given by p =-1r, +192n+18,'4where n is the number of television sets produced. Find the number of sets that must be- produced to maximise profit and calculaie the maximum profit.

5 Divide L50 into 2 non-zero integers r and y so that *y, i" a maximum.

/ +manufacturer makes open rylindrical containers, using 400 cm2 of aluminium sheeting.Determine the radius (to one decimal place), of the base"of the rylinder, so that the volumewill be a mafm,tT.Th: setting out foithis problem follows, buisome steps have not beencompleted. Complete the missing steps.

/ - 2nrh+ rrrz

V - ltrzh

400 - 2nrh+ 7Tr2 substitute in 1

2nfh1 ..............o......... .... a)

h-..............o....o.... .... b)

,200rftr2

V - ftrz x....,..........r.... .... c) substitute in 2

V - o...........o!..o....r '.... d)

form a rectangle. Determine the largestarea that can be formed with



Seni r

natima and minima problems - continaed

To determine where Vis a maximum, we find the derivative and 61(=Odr

dvdr

dTvdr2

Letdv-odr

Snrz = 400

i.e. 200-3nr2 - 02

{"'(t-t

o\...ataaao!..a...aao.. t o tt Yl

f) (since r>0maximum)

. d2vand -<0,V isadrz

r =............ ....g)

= 6.5 (to 1 decimal place)

For a maximum volume open rylinder the radius = 6'5 cm'

An aluminium can manufacturer wishes to make each can with a surface area of 600 cm2

If the cans have both a top and a bottom, determine (to 2 decimal places):

i) the radius of the can that will give the greatest volume

ii) the maximum volume of the can.

[Hint: Surface Area = 2nr2 +2nrh.l

g A manufacturer makes cartons (with no tops) that have square bases and rectangular sides.

Assume the base length is I cm and the height is h cm, as in the diagram. If the volume of

the box has to be 13 5=00 cm3, determine the dimensions of the box that will give the least

surface area.

A gardener creates an outdoor feature as shown below, which has an area of 98 cm2. The feature

coiprises a pond surrounded by grass. The grass strip is L metre wide along two sides.and.

0.5 metre wid.e along the other twJ sides. Deiermine the dimensions of the feature so that the

area of the pond is a maximum.

0.5 m

I

lmPond

@ Kevin Dunne and Blake Education 2006Senior Maths - Calculut



,/(

flNstvffs{'rfferentiation with rational powersl-1 1:2. 1-+ 2 -1 31i*, (2) i*t (3) .u*, @, i*t (s) i*,

d) 1*1 g) *+ (8) 2*t (e) a'* (10) lrt3

11)

-s*1 $2,)

-! *+ (13) Bx7 (14) T x6 (1s)-x-z

# $7r?*t (1s) # (1e)#

+ eD+ e2,+ (23)++ (2s) ,+ eo+ e7r+ (2s)

(16)

(20)

(24)

(2s)ry (30) + (31)# (s2) #(33)# Glo# (Bs)# (36)#

t: (38) ++8x (3e) 2x-+37' t* N* 2r x ' 3l:xn

(40)+-#(41)#-+ @4#-#2'lx

(4s) i @4):

8-differentiation - product rule(Il 6x2 +8x+4 (2) 3xz -10x+3 (3) 8x3 -9x2 +8

(41 3xz + 6x - L8 (5) 10ra + 36x3 + 4x + 9

(6) 15ra-36xs +27x2 -54x (71 4x3 -6x2 -8x+8(4 Lax6 +20xa -8x3 -8x (9) 7x6 -1,8x5 +24x3 -54x2

(10) lAxa -4x3 -24x2 +8x (11) 8x3 +75x2 +8(I2, 42x2 -3Ax-9 (13) Sxa +12x2 +\2x(14) 5xs + 4xs -3x2 -2x (15) 3x2 +I8x+\9(16) I}xa +I2x3 -l8xz -18-r (17) 96xs +708x2 -36x-27(18) 30xa +6x2 -8 (19) -30xa +32x3 -72x2 +64x(20) 150xa +760x3 -78x2 -76x (21,). 8x3 - t2xz +6x-5(221 48x3 -5'1.x2 +1,2x+4 (231 31 e4l 47 (25) 2

(26' -6 (271 4 (28) -3 (2e);

(30) 26x - y -48 = 0

9-differentiation r quotient rule

$)# e)# @)ffie)ffi

(s)ffi@#er# (8)#s& (10)#(11)#

1,

&

2x2 -8x+6 -x2 -3(r2' vr.Y (13) i(1s) 24x2 + 4ox (16) ,

u*\ z

(6x+f\--'

@'-I)'

(r4)ffi2xa +3xz(I7)W

(18) ffi (1e) + (20) -i (2r')7" (22r 42"

(23) x - y+ 5 - 0 (24) \lx + Y - L3 = 0

(25)8x+9y+7=0(26)3x-4v_1=0,8x+5y_11=0

l0-diffetentiation .' chain rule(1) 12( 4x+2)2 (2' 20(5x+3)3 (3) -24(7 -6x)'

(4) 35(7x-,3)a (5) 6x(x2 + 4)' (61 zhc@*-Z)z

(7, -36x2(2-3x')' (8) 50r3(3x++$)+ (9) -I4x(7x'-3)-

(1o) -i(ix-))-' (11) & $2) ffi2

(13)

W$4, (1s)

(1o) -14(7x-3)* (l:7') -ift-*l'i (18) L72

/-3

t ';-' xkz.'- tr(1e) @ (2o)

@(zrt@ Q2l-72

(23) 876204 (241 + eil s eq'n,-*,

-ll-calculus review 2

0 + (2) 3x2 +6x-2 @) # @, 6(s+zx)2

(s) # (6) ffi (7)4xj+zx (8)#(e) -16(3 -4x)' (1o)

&(11) i0xa-6xz+L2x

$u+ (13) ffi sq -r6x(r-4x')

(1s) # (16) + $n sxa -4xz +Zx-I

(18)&

(1e) x(x,+3)* (-r3x3 -ex+30)

(2o) ffi eD + e2l 4(3xz+3)(x3 +3x+4),

(2s, ffi Qq (x' 1)(8xa +21x3 -2x-3)

(2s) ffi=# (z6)ffi en*,tt(2s)

=,ry Qs, 2x- y - 4=a (30) 108

25',4

l3-Iocal (relative) maxima and minimaand hofizontal points of inflexionusing the first derivative

(1) max (2) inflexion (3) rnax (a) inflexion (5) min(6) inflexion (7) inflexion (B)

min(9) min

(10) inflexion (fSl rnax, (A,-6) (16) min, (-4,-IL)/

(17) ma&li,^)

(18) milr (-!,4); min, (r,- 4)

(19) max, ('-4,28);min, (-7,1) (20) inflexion (2:rl (;,t)' -r, r:),miru (3,-5) (2s)max, (z,e)221 *"rI

(24, min, (-1,-7) (25) mali (-I,Z); mrn, (I,-2)(26) max, (-3,75); min, (0,48)

2O-maxima and minima problems(l) 36 m2 (2) 1250m2 (g) 'J,6,1,6 (4) I28, $12303

(5) 100,50 (6) (a) 400-TErz p,1 Wznr (c)

(d) zoo,-+ (e) zoo -T (r) -3tcr ,r, E(7') 5.64 cm,1128.38 cm2 (g) 30 cffi, 30 cm, L5 cm(9) 1,4 m,7 rn

200 _LIcr 2

Differentiation Revision Worksheets

Documents

x f x 6x3

xs v x

xxt x

x 3x22x

4x2xz x

u f x

given f x

derivative of f x