worksheet 4.1 derivatives of exponential and logarithmic
TRANSCRIPT
Maths Quest Maths B Year 12 for Queensland 2e 1
WorkSHEET 4.1 Derivatives of exponential and logarithmic functions Name: ___________________________ 1 As an example for the External test β¦ what we
know and do already can look completely different. When we do Inverse functions, we swap x and y and then rearrange. However in this question it asks to simply rearrange to get x= . Then as a part b) it says to state the inverse function. It looks different, but it is just what we have always done, just βdressed differentlyβ. If : (a) express x in terms of y.
(b) find the equation of the inverse.
You can also see the Textbook Solution doesnβt state the base it uses, but it would seem common sense to choose Base 3 to make the logs disappear? Iβd prefer an answer:
π¦!" =log# π¦ β 1
3
2 If : (c) express x in terms of y.
(d) find the equation of the inverse. This is another Textbook question and solution β¦ Iβll leave it here as itβs good practice, but donβt get too stressed about how you leave your answer as there are a number of different correct forms!
133 += xy
( )
( )( )3log3
3loglog3log33loglog
3log3log3log3log13log
3 (a) 13
-=
=-+=
+== +
yx
xyxyxy
y x
( )( )3log3
3loglog (b) -=
xy
y = e3x+1 (a) y = 33x+1
ln y = 3x +1( ) lneln y = 3x +1
log yβ1= 3x
x =ln yβ1( )
3
(b) y =ln x β1( )
3
Maths Quest Maths B Year 12 for Queensland 2e 2
3 Find the inverse of the function;
π(π₯) = π$%&'
Convert to familiar terminology;
π¦ = π$%&' Swap π₯ππππ¦.
π₯ = π$(&'
ln π₯ = ln π$(&'
2π¦ + 4 = ln π₯
π¦ =ln π₯ β 4
2 Try and go back to put the !" in your work
β΄ π¦!" =ln π₯ β 4
2
4 Express y in terms of x in :
yx ee log12log4 =-
( )( )
( )
( )
( )
( ) yex
yex
eyx
e
y
x
y
x
yx
yx
e
ee
ee
=
=
=
=
=
=-
=-
2
8
214
21
4
1
21
4
21
4
21
4
2
2
2
2
12log
1log2log
log12log4
Maths Quest Maths B Year 12 for Queensland 2e 3
5 Differentiate, showing your FULL use of the Chain rule: NEW Curriculum. This is a Tech FREE
question, hence no need for full working
Set,
π = ππ β π ππ π = ππ
Where,
π = πππ β π ππ π = ππ
Use Chain rule;
π ππ π =
π ππ π Γ
π ππ π
= ππ Γ ππ
= ππππππ
6 NEW Curriculum. This Q would be in a Tech FREE exam, hence no need for full working! Differentiate:
π π ππ
πππ = ππππππ Is Full Marks! (itβs only a 1 mark question L )
7 Differentiate:
ππ
π π ππ
π =ππ
8 Differentiate each of the following:
Differentiate by βinspectionβ:
βππ!ππ
9 Differentiate each of the following: (a)
(b)
22xe
22xe
xe 22 -
xe 22 -
22xe
2
2
(a) Let 2d 4d
x
x
y ey ex
-
-
=
= -
2
2
2
2
(b) Let d 4d
x
x
y ey xex
=
=
Maths Quest Maths B Year 12 for Queensland 2e 4
10 Differentiate each of the following: (a) (b)
If itβs HARD, it may be best to actually use the Chain Rule? Set,
π = ππ β π ππ π = ππ
Where,
π = πππ β
π ππ π =
πππ
!ππ =ππβπ
Use Chain Rule;
π ππ π =
π ππ π
π ππ π
= ππππβπ
=πβπ
πβπ
Always SIMPLIFY first β¦ Expand brackets! Set,
π = πππ β π Now by inspection,
π/ = ππππ
xe
( )xxx eee --3
Maths Quest Maths B Year 12 for Queensland 2e 5
11 Differentiate each of the following: (a) s
(b)
(a) π¦ = (3π!% β π%)$
= 9π!$% β 6π!%π% + π$%
β΄ π¦ = π$% + 9π!$% β 6
Now,
ππ¦ππ₯ = 2π$% β
18π$%
ππ¦ππ₯ = π% β
2π%
12 Differentiate each of the following: (a) (b)
( ) 23 xx ee --
x
xx
eee
2
3 2+3
22(b) Let
2d 2d
x x
x
x x
x x
e eye
y e ey e ex
-
-
+=
= +
= -
xe 4log
)41(log4 xe -
(a) Let log 4 4d 1 dlog 4d d
d d dd d d
1 4
1 441
e
e
y x u xy uy uu u x
y y ux u x
u
x
x
= =
= = =
= Β΄
= Β΄
= Β΄
=
(b) Let 4 log (1 4 ) 1 4d 4 d4log 4d d
d d dd d d
4 4
161 4
e
e
y x u xy uy uu u x
y y ux u x
u
x
= - = -
= = = -
= Β΄
= Β΄-
= --
Maths Quest Maths B Year 12 for Queensland 2e 6
13 Differentiating a Logarithm is just a pattern. It comes from using the Chain Rule; Differentiate:
π¦ = ln(π₯$ + 2π₯ + 6)
Set,
π = π₯π§π β π ππ π =
ππ
Where,
π = ππ + ππ + π β π ππ π = ππ + π
Use Chain Rule;
π ππ π =
π ππ π
π ππ π
=ππ Γ (ππ + π)
π ππ π =
π(π + π)ππ + ππ + π
14
Differentiate:
π¦ = ln(π₯$ + 3π₯ + 5)
Without the working.
ππππ πππππππππππππππππππππππππππππππ
no, donβt quote that rule in the exam!
π/ =ππ + π
ππ + ππ + π The rule looks like:
π π π π₯π§Wπ
(π)X =πβ²(π)π(π)
15 Some simple ones, Differentiate:
ln π₯
ln 5π₯
ln(π₯ + 1)
ln(2π₯ + 5)
Some simple ones, Differentiate:
πππ₯ ln π₯ =
1π₯
πππ₯ ln 5π₯ =
55π₯ =
1π₯
πππ₯ ln
(2π₯ + 5) =2
2π₯ + 5
Maths Quest Maths B Year 12 for Queensland 2e 7
16 Some simple ones, Differentiate: ** Make sure you SIMPLIFY your answer!
ln(2π₯ + 4)
ln(5π₯ + 10)
ln(6π₯ + 42)
Some simple ones, Differentiate:
πππ₯ ln
(2π₯ + 4) =2
2π₯ + 4 =2
2(π₯ + 2) =1
π₯ + 2
πππ₯ ln(5π₯ + 10) =
55π₯ + 10 =
1π₯ + 2
πππ₯ ln(6π₯ + 42) =
1π₯ + 7
17 Differentiate each of the following:
(a)
(b)
Γ·ΓΈΓΆ
çèæ
+141logxe
)4(log 3 xxe -
( )
( )
11(a) Let log log 4 14 1
log 4 14 1
d 1 dlog 4d d
d d dd d d
1 4
44 1
e e
e
e
y xxx
u xy uy uu u x
y y ux u x
u
x
-= = ++
= - +
= +
= - = - =
= Β΄
= - Β΄
= -+
( )
3 3
2
2
2
3
(b) Let log (4 ) 4d 1 dlog 12 1d d
d d dd d d
1 12 1
12 14
e
e
y x x u x xy uy u xu u x
y y ux u x
xuxx x
= - = -
= = = -
= Β΄
= Β΄ -
-=
-
Maths Quest Maths B Year 12 for Queensland 2e 8
18 Differentiate each of the following: (a)
(b)
22log4 xe -
( )531logx
e-
( )( )
2 2
12 2
2
2
2
(a) Let 4 log 2 2
4log 2
2log 2
d 2 d2log 2d d
d d dd d d
2 2
424
2
e
e
e
e
y x u x
x
x
y uy u xu u x
y y ux u x
xu
xx
xx
= - = -
-
= -
= = = -
= Β΄
= Β΄-
= --
=-
( )( )( )
5
5
1(b) Let log 33
log 3
5log 3d 5 d5log 1d d
d d dd d d
5 1
53
e
e
e
e
y u xx
y x
y xy uy uu u x
y y ux u x
u
x
-
= = --
= -
= - -
= - = - = -
= Β΄
= - Β΄-
=-
Maths Quest Maths B Year 12 for Queensland 2e 9
19 Find exact values of if =:
(a)
20 Find exact values of if
=
Eeeek β¦ Yukky! Here, let me show you easy a product rule is with my setting out and giving your eyes some space to see whats going on!
21 Refer to last question, but now the solution setting out is much βnicerβ The whole point of manually differentiating is to present your answer in FACTOR FORM β¦ !!! Factor form could be worth a mark in your test, so make sure you fctor your derivaitive questions where possible!
π¦ = 8π#%(4π₯# β 3) Set,
π¦ = π’ Γ π£ Where,
π’ = 8π#% β π’/ = 24π#% and
π£ = 4π₯# β 3 β π£/ = 12π₯$ Use Product rule;
π¦/ = π£π’/ + π’π£/
= (4π₯# β 3) Γ 24π#% + 8π#% Γ 12π₯$
= 24π#%(4π₯# β 3) + 96π₯$π#%
= 96π₯#π#% β 96π#% + 72π₯$π#%
β΄ πβ²(π₯) = π#%(96π₯# + 96π₯$ β 72) Now,
π/(β1) = β72π!# = β72π#
( )3f Β’ ( )xf
216log2 xe --
( )
( ) ( )( ) ( )
( ) ( )
( ) ( )
( )
2
12 2
2 2
2
2
(b) 2 log 16
2log 16
log 16 16
d 1 dlog 2d d
d d dd d d
1 2
216
2 3 6316 3 7
e
e
e
e
f x x
f x x
f x x u x
f x uf x u xu u x
f x f x ux u x
xuxx
f
= - -
= - -
= - - = -
= - = - = -
= Β΄
= - Β΄-
=-´’ = =-
( )1-Β’f( )xf ( )348 33 -xe x
( )
( ) ( )( )
3 3 3 3
3 2
3 2 3 3
3 3 3
(a) Let 8 4 3 , 8 , 4 3
d d d d d, 24 , 12d d d d d
8 12 4 3 24
1 96 168 72
x x
x
x x
y e x u e v x
y v u u vu v e xx x x x x
f x e x x e
f e e e- - -
= - = = -
= + = =
= Β΄ + - Β΄
- = - = -
Maths Quest Maths B Year 12 for Queensland 2e 10
22 Differentiate;
π¦ = π% ln π₯
π¦ = π% ln π₯ Set,
π¦ = π’ Γ π£ Where,
π’ = π% β π’/ = π% and
π£ = ln π₯ β π£/ =1π₯
Use Product rule; π¦/ = π£π’/ + π’π£/
= ln π₯ Γ π% + π% Γ1π₯
= π% ln π₯ +π%
π₯
= π% ^ln π₯ +1π₯_
23 Differentiate;
π¦ = π#% ln 4π₯
π¦ = π#% ln 4π₯ Set,
π¦ = π’ Γ π£ Where,
π’ = π#% β π’/ = 3π#% and
π£ = ln 4π₯ β π£/ =1π₯
Use Product rule; π¦/ = π£π’/ + π’π£/
= ln 4π₯ Γ 3π#% + π#% Γ1π₯
= 3π#% ln 4π₯ +π#%
π₯
= π% ^3 ln 4π₯ +1π₯_
Maths Quest Maths B Year 12 for Queensland 2e 11
24 Differentiate;
π¦ = π% ln π₯$
First thing is first, simplify!
π¦ = 2π% ln π₯ Set,
π¦ = π’ Γ π£ Where,
π’ = 2π% β π’/ = 2π% and
π£ = ln π₯ β π£/ =1π₯
Use Product rule; π¦/ = π£π’/ + π’π£/
= ln π₯ Γ 2π% + 2π% Γ1π₯
= 2π% ln π₯ +2π%
π₯
= 2π% ^ln π₯ +1π₯_
Maths Quest Maths B Year 12 for Queensland 2e 12
25 Determine:
πππ₯ln(3π₯ + 2) π!"#{%#('()*)}
Product rule, set:
π¦ = π’ Γ π£ where;
π’ = ln(3π₯ + 2) ππππ£ = π012{42(#%&$)}
π’ = lnπ€ β ππ’ππ€ =
1π€
π€ = 3π₯ + 2 β ππ€ππ₯ = 3
π’/ =ππ’ππ₯ =
ππ¦ππ€ Γ
ππ€ππ₯
=1π€ Γ 3
β΄ π’β² =3
3π₯ + 2
πππ
π£ = π8 β ππ£ππ§ = π8
π§ = sin π’ β ππ§ππ’ = cos π’
π’ = ln(3π₯ + 2) β ππ’ππ₯ =
33π₯ + 2
*** note: Did you see I set π§ = sin π’ β¦ usually I would have used a different variable (I have already used π’) , but in this case, I have already determined the derivative of ln(3π₯ + 2), so there is no need to do it again, so we can use the previous working to skip all that and go directly to β¦ 9:
9%= #
#%&$ ***
π£/ =ππ£ππ₯ =
ππ£ππ§ Γ
ππ§ππ’ Γ
ππ’ππ₯
= π8 cos π’3
3π₯ + 2
β΄ π£/ =3 cos(ln(3π₯ + 2)π012{42(#%&$)}
3π₯ + 2
Maths Quest Maths B Year 12 for Queensland 2e 13
26 Yep, this solution took 2 pages β¦ J β¦ and now I am doubling the width of the solutions page β¦ looks like fun hey! Contβd Now,
π¦/ = π£π’/ + π’π£/ so,
π¦- = π!"#{%#('()*)} Γ3
3π₯ + 2+ ln(3π₯ + 2) Γ
3π!"#{%#('()*)} cos{ln(3π₯ + 2)}3π₯ + 2
π¦- =3π!"#{%#('()*)}
3π₯ + 2+3π!"#{%#('()*)} cos{ln(3π₯ + 2)} ln(3π₯ + 2)
3π₯ + 2
π¦- =3π!"#{%#('()*)}
3π₯ + 2(1 + ln(3π₯ + 2)cos {ln(3π₯ + 2)})