WorkSHEET 4.1 Derivatives of exponential and logarithmic

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 … 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 “dressed differently”. If : (a) express x in terms of y.

(b) find the equation of the inverse.

You can also see the Textbook Solution doesn’t state the base it uses, but it would seem common sense to choose Base 3 to make the logs disappear? I’d prefer an answer:

𝑦!" =log# 𝑦 − 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 … I’ll leave it here as it’s good practice, but don’t 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 y−1= 3x

x =ln y−1( )

3

(b) y =ln x −1( )

3


3 Find the inverse of the function;

𝑓(𝑥) = 𝑒$%&'

Convert to familiar terminology;

𝑦 = 𝑒$%&' Swap 𝑥𝑎𝑛𝑑𝑦.

𝑥 = 𝑒$(&'

ln 𝑥 = ln 𝑒$(&'

2𝑦 + 4 = ln 𝑥

𝑦 =ln 𝑥 − 4

2 Try and go back to put the !" in your work

∴ 𝑦!" =ln 𝑥 − 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


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,

𝒚 = 𝒆𝒖 → 𝒅𝒚𝒅𝒖 = 𝒆𝒖

Where,

𝒖 = 𝟐𝒙𝟐 → 𝒅𝒖𝒅𝒙 = 𝟒𝒙

Use Chain rule;

𝒅𝒚𝒅𝒙 =

𝒅𝒚𝒅𝒖 ×

𝒅𝒖𝒅𝒙

= 𝒆𝒖 × 𝟒𝒙

= 𝟒𝒙𝒆𝟐𝒙𝟐

6 NEW Curriculum. This Q would be in a Tech FREE exam, hence no need for full working! Differentiate:

𝒅𝒅𝒙𝒆

𝟐𝒙𝟐 = 𝟒𝒙𝒆𝟐𝒙𝟐 Is Full Marks! (it’s only a 1 mark question L )

7 Differentiate:

𝒆𝒙

𝒅𝒅𝒙𝒆

𝒙 =𝒆𝒙

8 Differentiate each of the following:

Differentiate by “inspection”:

−𝟒𝒆!𝟐𝒙

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

=

=


10 Differentiate each of the following: (a) (b)

If it’s HARD, it may be best to actually use the Chain Rule? Set,

𝒚 = 𝒆𝒖 → 𝒅𝒚𝒅𝒖 = 𝒆𝒖

Where,

𝒖 = 𝒙𝟏𝟐 →

𝒅𝒖𝒅𝒙 =

𝟏𝟐𝒙

!𝟏𝟐 =𝟏𝟐√𝒙

Use Chain Rule;

𝒅𝒚𝒅𝒙 =

𝒅𝒚𝒅𝒖

𝒅𝒖𝒅𝒙

= 𝒆𝒖𝟏𝟐√𝒙

=𝒆√𝒙

𝟐√𝒙

Always SIMPLIFY first … Expand brackets! Set,

𝒚 = 𝒆𝟒𝒙 − 𝟏 Now by inspection,

𝒚/ = 𝟒𝒆𝟒𝒙

xe

( )xxx eee --3


11 Differentiate each of the following: (a) s

(b)

(a) 𝑦 = (3𝑒!% − 𝑒%)$

= 9𝑒!$% − 6𝑒!%𝑒% + 𝑒$%

∴ 𝑦 = 𝑒$% + 9𝑒!$% − 6

Now,

𝑑𝑦𝑑𝑥 = 2𝑒$% −

18𝑒$%

𝑑𝑦𝑑𝑥 = 𝑒% −

2𝑒%

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

= =

= = =

= ´

= ´

= ´

=

(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

= - = -

= = = -

= ´

= ´-

= --


13 Differentiating a Logarithm is just a pattern. It comes from using the Chain Rule; Differentiate:

𝑦 = ln(𝑥$ + 2𝑥 + 6)

Set,

𝒚 = 𝐥𝐧𝒖 → 𝒅𝒚𝒅𝒖 =

𝟏𝒖

Where,

𝒖 = 𝒙𝟐 + 𝟐𝒙 + 𝟔 → 𝒅𝒖𝒅𝒙 = 𝟐𝒙 + 𝟐

Use Chain Rule;

𝒅𝒚𝒅𝒙 =

𝒅𝒚𝒅𝒖

𝒅𝒖𝒅𝒙

=𝟏𝒖 × (𝟐𝒙 + 𝟐)

𝒅𝒚𝒅𝒙 =

𝟐(𝒙 + 𝟏)𝒙𝟐 + 𝟐𝒙 + 𝟔

14

Differentiate:

𝑦 = ln(𝑥$ + 3𝑥 + 5)

Without the working.

𝒕𝒉𝒆𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆𝒐𝒇𝒕𝒉𝒆𝒃𝒓𝒂𝒄𝒌𝒆𝒕𝒕𝒉𝒆𝒃𝒓𝒂𝒄𝒌𝒆𝒕

no, don’t quote that rule in the exam!

𝒚/ =𝟐𝒙 + 𝟑

𝒙𝟐 + 𝟑𝒙 + 𝟓 The rule looks like:

𝒅𝒅𝒙 𝐥𝐧W𝒇

(𝒙)X =𝒇′(𝒙)𝒇(𝒙)

15 Some simple ones, Differentiate:

ln 𝑥

ln 5𝑥

ln(𝑥 + 1)

ln(2𝑥 + 5)

Some simple ones, Differentiate:

𝑑𝑑𝑥 ln 𝑥 =

1𝑥

𝑑𝑑𝑥 ln 5𝑥 =

55𝑥 =

1𝑥

𝑑𝑑𝑥 ln

(2𝑥 + 5) =2

2𝑥 + 5


16 Some simple ones, Differentiate: ** Make sure you SIMPLIFY your answer!

ln(2𝑥 + 4)

ln(5𝑥 + 10)

ln(6𝑥 + 42)

Some simple ones, Differentiate:

𝑑𝑑𝑥 ln

(2𝑥 + 4) =2

2𝑥 + 4 =2

2(𝑥 + 2) =1

𝑥 + 2

𝑑𝑑𝑥 ln(5𝑥 + 10) =

55𝑥 + 10 =

1𝑥 + 2

𝑑𝑑𝑥 ln(6𝑥 + 42) =

1𝑥 + 7

17 Differentiate each of the following:

(a)

(b)

÷øö

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

-= = ++

= - +

= +

= - = - =

= ´

= - ´

= -+

( )

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

= - = -

= = = -

= ´

= ´ -

-=

-


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

= - = -

-

= -

= = = -

= ´

= ´-

= --

=-

( )( )( )

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

-

= = --

= -

= - -

= - = - = -

= ´

= - ´-

=-


19 Find exact values of if =:

(a)

20 Find exact values of if

=

Eeeek … 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 “nicer” The whole point of manually differentiating is to present your answer in FACTOR FORM … !!! Factor form could be worth a mark in your test, so make sure you fctor your derivaitive questions where possible!

𝑦 = 8𝑒#%(4𝑥# − 3) Set,

𝑦 = 𝑢 × 𝑣 Where,

𝑢 = 8𝑒#% → 𝑢/ = 24𝑒#% and

𝑣 = 4𝑥# − 3 → 𝑣/ = 12𝑥$ Use Product rule;

𝑦/ = 𝑣𝑢/ + 𝑢𝑣/

= (4𝑥# − 3) × 24𝑒#% + 8𝑒#% × 12𝑥$

= 24𝑒#%(4𝑥# − 3) + 96𝑥$𝑒#%

= 96𝑥#𝑒#% − 96𝑒#% + 72𝑥$𝑒#%

∴ 𝑓′(𝑥) = 𝑒#%(96𝑥# + 96𝑥$ − 72) Now,

𝑓/(−1) = −72𝑒!# = −72𝑒#

( )3f ¢ ( )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

= - -

= - -

= - - = -

= - = - = -

= ´

= - ´-

=-´¢ = =-

( )1-¢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- - -

= - = = -

= + = =

= ´ + - ´

- = - = -


22 Differentiate;

𝑦 = 𝑒% ln 𝑥

𝑦 = 𝑒% ln 𝑥 Set,


𝑢 = 𝑒% → 𝑢/ = 𝑒% and

𝑣 = ln 𝑥 → 𝑣/ =1𝑥

Use Product rule; 𝑦/ = 𝑣𝑢/ + 𝑢𝑣/

= ln 𝑥 × 𝑒% + 𝑒% ×1𝑥

= 𝑒% ln 𝑥 +𝑒%

𝑥

= 𝑒% ^ln 𝑥 +1𝑥_

23 Differentiate;

𝑦 = 𝑒#% ln 4𝑥

𝑦 = 𝑒#% ln 4𝑥 Set,


𝑢 = 𝑒#% → 𝑢/ = 3𝑒#% and

𝑣 = ln 4𝑥 → 𝑣/ =1𝑥


= ln 4𝑥 × 3𝑒#% + 𝑒#% ×1𝑥

= 3𝑒#% ln 4𝑥 +𝑒#%

𝑥

= 𝑒% ^3 ln 4𝑥 +1𝑥_


24 Differentiate;

𝑦 = 𝑒% ln 𝑥$

First thing is first, simplify!

𝑦 = 2𝑒% ln 𝑥 Set,


𝑢 = 2𝑒% → 𝑢/ = 2𝑒% and

𝑣 = ln 𝑥 → 𝑣/ =1𝑥


= ln 𝑥 × 2𝑒% + 2𝑒% ×1𝑥

= 2𝑒% ln 𝑥 +2𝑒%

𝑥

= 2𝑒% ^ln 𝑥 +1𝑥_


25 Determine:

𝑑𝑑𝑥ln(3𝑥 + 2) 𝑒!"#{%#('()*)}

Product rule, set:

𝑦 = 𝑢 × 𝑣 where;

𝑢 = ln(3𝑥 + 2) 𝑎𝑛𝑑𝑣 = 𝑒012{42(#%&$)}

𝑢 = ln𝑤 → 𝑑𝑢𝑑𝑤 =

1𝑤

𝑤 = 3𝑥 + 2 → 𝑑𝑤𝑑𝑥 = 3

𝑢/ =𝑑𝑢𝑑𝑥 =

𝑑𝑦𝑑𝑤 ×

𝑑𝑤𝑑𝑥

=1𝑤 × 3

∴ 𝑢′ =3

3𝑥 + 2

𝑎𝑛𝑑

𝑣 = 𝑒8 → 𝑑𝑣𝑑𝑧 = 𝑒8

𝑧 = sin 𝑢 → 𝑑𝑧𝑑𝑢 = cos 𝑢

𝑢 = ln(3𝑥 + 2) → 𝑑𝑢𝑑𝑥 =

33𝑥 + 2

*** note: Did you see I set 𝑧 = sin 𝑢 … usually I would have used a different variable (I have already used 𝑢) , but in this case, I have already determined the derivative of ln(3𝑥 + 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 … 9:

9%= #

#%&$ ***

𝑣/ =𝑑𝑣𝑑𝑥 =

𝑑𝑣𝑑𝑧 ×

𝑑𝑧𝑑𝑢 ×

𝑑𝑢𝑑𝑥

= 𝑒8 cos 𝑢3

3𝑥 + 2

∴ 𝑣/ =3 cos(ln(3𝑥 + 2)𝑒012{42(#%&$)}

3𝑥 + 2


26 Yep, this solution took 2 pages … J … and now I am doubling the width of the solutions page … looks like fun hey! Cont’d Now,

𝑦/ = 𝑣𝑢/ + 𝑢𝑣/ so,

𝑦- = 𝑒!"#{%#('()*)} ×3

3𝑥 + 2+ ln(3𝑥 + 2) ×

3𝑒!"#{%#('()*)} cos{ln(3𝑥 + 2)}3𝑥 + 2

𝑦- =3𝑒!"#{%#('()*)}

3𝑥 + 2+3𝑒!"#{%#('()*)} cos{ln(3𝑥 + 2)} ln(3𝑥 + 2)

3𝑥 + 2

𝑦- =3𝑒!"#{%#('()*)}

3𝑥 + 2(1 + ln(3𝑥 + 2)cos {ln(3𝑥 + 2)})

Documents

WorkSHEET 4.1 Derivatives of exponential and logarithmic