Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
23
Notes
DIFFERENTIATION OF EXPONENTIAL
AND LOGARITHMIC FUNCTIONS
We are aware that population generally grows but in some cases decay also. There are many
other areas where growth and decay are continuous in nature. Examples from the fields of
Economics, Agriculture and Business can be cited, where growth and decay are continuous.
Let us consider an example of bacteria growth. If there are 10,00,000 bacteria at present and
say they are doubled in number after 10 hours, we are interested in knowing as to after how
much time these bacteria will be 30,00,000 in number and so on.
Answers to the growth problem does not come from addition (repeated or otherwise), or
multiplication by a fixed number. In fact Mathematics has a tool known as exponential function
that helps us to find growth and decay in such cases. Exponential function is inverse of logarithmic
function.
In this lesson, we propose to work with this tool and find the rules governing their derivatives.
OBJECTIVES
After studying this lesson, you will be able to :
●
define and find the derivatives of exponential and logarithmic functions;
●
find the derivatives of functions expressed as a combination of algebraic, trigonometric,
exponential and logarithmic functions; and
●
find second order derivative of a function.
EXPECTED BACKGROUND KNOWLEDGE
l
Application of the following standard limits :
n
(i)
(iii)
(v)
l
1

lim 1 +  = e
n→∞ 
n
x
e −1
lim
=1
x→∞ x
lim
h→ 0
(eh − 1)
h
(ii)
(iv)
lim (1+ n )
n→∞
1
n
=e
a x −1
= log e a
x→∞ x
lim
=1
Definition of derivative and rules for finding derivatives of functions.
MATHEMATICS
277
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
23.1 DERIVATIVE OF EXPONENTIAL FUNCTIONS
Let y = e x be an exponential function.
.....(i)
y + δy = e( x +δx ) (Corresponding small increments)
∴
.....(ii)
From (i) and (ii), we have
Notes
δy = e x +δ x − e x
∴
Dividing both sides by δx and taking the limit as δx → 0
∴
[e δ x − 1]
δy
= lim e x
δ x→ 0 δ x δ x→ 0
δx
⇒
dy
= e x .1 =e x
dx
Thus, we have
Working rule :
Next, let
Then
lim
( )
d x
e = ex .
dx
( )
d x
d
e = ex ⋅ ( x ) = ex
dx
dx
y = e ax+ b .
y + δy =ea(x + δx ) +b
[ δx and δy are corresponding small increments]
∴
δ y = ea
(x + δx ) +b
−e ax
+b
= eax +b e a δx −1


e aδ x − 1
δy
ax +b 

=e
δx
δx
∴
= a ⋅e ax + b
ea δx −1
aδ x
[Multiply and divide by a]
Taking limit as δ x → 0 , we have
lim
δ x →0
or
δy
eaδ x − 1
= a ⋅ eax + b ⋅ lim
δx
δ x →0 a δx
dy
= a ⋅ e ax + b ⋅1
dx

ex − 1 
= 1
 lim
 x →0 x

= aeax + b
278
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
Working rule :
d ( ax +b ) ax +b d
e
=e
⋅ (ax+ b ) = eax
dx
dx
+b
⋅a
∴
d ( a x+ b )
e
= ae ax + b
dx
Example 23.1
Find the derivative of each of the follwoing functions :
(i) e
5x
(ii) e
ax
(iii) e
−
MODULE - V
Calculus
Notes
3x
2
Soution : (i) Let y = e5 x .
y = e t where 5 x = t
Then
dy
= et
dt
∴
and 5 =
dt
dx
We know that, dy = dy ⋅ dt = et ⋅ 5 = 5e5x
dx
dt dx
( )
Alternatively
d 5x
d
5x
e
= e5x ⋅ ( 5x ) = e5x ⋅ 5 = 5e
dx
dx
(ii) Let
y = eax ⋅
Then
y = e t when t = ax
∴
dy
dt
= et and
=a
dt
dx
We know that,
dy dy dt
=
×
= et ⋅ a
dx dt dx
Thus,
dy
= a ⋅ eax
dx
−3x
=e 2
(iii) Let
y
∴
dy
=e
dt
dy −3
= e
dt
2
Thus,
Example 23.2
(i)
−3
x
2
−3x
2
Find the derivative of each of the following :
y = e x + 2cosx
MATHEMATICS
d  −3 
⋅ 
x
dx  2 
2
5
3
(ii) y = ex + 2sinx − e x + 2e
279
Differentiation of Exponential and Logarithmic Functions
MODULE - V Solution : (i)
Calculus
y = e x + 2cosx
dy d ( x )
d
=
e +2
( c o s x ) = ex − 2sinx
dx dx
dx
∴
2
5
y = ex + 2sinx − e x + 2e
3
(ii)
( )
Notes
2 d
dy
5
= ex
x 2 +2 c o s x − ex +0 [Since e is constant]
dx
dx
3
∴
2
5
= 2xex + 2cosx − e x
3
Example 23.3 Find
(i)
y=e
dy
, when
dx
xcosx
Solution : (i)
(ii)
y=
1 x
e
x
(iii)
y=
1− x
e 1+ x
y = e xcosx
∴
dy
d
= ex c o s x
(x c o s x )
dx
dx
∴
dy
d

xcosx  d
=e
x
cosx + cosx
(x) 

dx
dx
 dx

= excosx [ −xsinx + cosx]
(ii)
∴
y=
( )
dy
x d 1  1 d
x
=e
e
 +
dx
dx  x  x dx
=
=
(iii)
1 x
e
x
y=
−1
x
2
ex +
[Using product rule]
1 x
e
x
ex
ex
x
x2
[−1 + x] =
2
[x − 1]
1− x
e1+ x
1− x
dy
d 1 − x 
= e 1+ x
dx
dx  1+ x 
280
=
1− x
e1+ x
 −1.(1+ x) − (1 − x).1 



(1 + x)2

=
1− x
e1+ x
 −2 
−2
=
e 1+ x

2
 (1 + x)  (1 + x)2
1− x
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
Example 23.4
(i)
MODULE - V
Calculus
Find the derivative of each of the following functions :
(ii) e ax ⋅ cos ( bx + c )
esinx ⋅ sinex
y = esinx ⋅ sinex
Solution :
(
)
dy
d
d
= esinx ⋅
sine x + sinex esinx
dx
dx
dx
∴
= esinx ⋅cose x ⋅
Notes
( )
d x
d
e + sinex ⋅ esinx ( sinx )
dx
dx
= esinx ⋅cose x ⋅ ex + sinex ⋅ esinx ⋅ cosx
= esinx [e x ⋅ cose x + sine x ⋅ cosx]
y = eax cos(bx + c)
(ii)
dy
d
d
= eax ⋅ cos(bx + c) + cos ( bx + c) eax
dx
dx
dx
∴
= eax ⋅ [− sin(bx + c)]
d
d
(bx + c) + cos(bx + c)eax
(ax)
dx
dx
= −e ax sin(bx + c)⋅ b + cos(bx + c)e ax .a
= eax [−bsin(bx + c) + acos(bx + c)]
Example 23.5 Find
dy
, if
dx
dy
=
dx
Solution :
y=
sin(bx + c)
=
=
eax
sin(bx + c)
d ax
d
e − eax [sin(bx + c)]
dx
dx
2
sin (bx + c)
sin(bx + c).e ax .a − e ax cos(bx + c).b
sin 2 (bx + c)
eax [asin(bx + c) − bcos(bx + c)]
sin 2 (bx + c)
CHECK YOUR PROGRESS 23.1
1. Find the derivative of each of the following functions :
(a) e
2.
5x
Find
(b) e
7x+ 4
(c) e 2x
(d)
−7
x
e2
(e) e x
2 + 2x
dy
, if
dx
MATHEMATICS
281
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
1
3
3.
(b) y = tanx + 2sinx + 3cosx − e x
(c) y = 5sinx − 2e x
(d) y = e x + e − x
Find the derivative of each of the following functions :
(a) f(x) = e
Notes
x +1
(c) f(x) = exsin
4.
1
2
(a) y = e x − 5e
(b) f(x) = e
2
2
(d) f(x) = exsec
x
x
Find the derivative of each of the following functions :
(a) f(x) = (x − 1)ex
5.
cotx
Find
(b) f(x) = e2 x sin 2 x
dy
, if
dx
(a) y =
e2x
e 2x ⋅ c o s x
(b) y =
x sin x
x2 +1
23.2 DERIVATIVE OF LOGARITHMIC FUNCTIONS
We first consider logarithmic function
Let
y = logx
∴
.....(i)
y + δ y = log ( x + δx )
.....(ii)
( δ x and δ y are corresponding small increments in x and y)
From (i) and (ii), we get
δ y = log (x + δx ) −l o g x
= log
∴
x + δx
x
δy 1
 δx 
=
log 1 + 
δx δx
x

=
1 x
 δx 
⋅ log 1+
x δx
x 

[Multiply and divide by x]
x
1
 δ x  δx
= log 1 + 
x
x

Taking limits of both sides, as δ x → 0 , we get
x
δy 1
 δ x  δx
lim
=
lim log 1 + 
δ x→ 0 δx
x δ x→ 0
x 

282
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
x 

dy 1

 δx  δ x 
= ⋅ log  lim  1 +  
dx x
x  
δ x→0 



Q lim
 δ x→0
1
= loge
x
=
x

δ
x
δ
1 +  x =e 




x 
Notes
1
x
d
1
(logx) =
dx
x
Next, we consider logarithmic function
Thus,
y = log(ax + b)
∴
y + δy = log[a(x + δx) + b]
...(i)
...(ii)
[ δx and δy are corresponding small increments]
From (i) and (ii), we get
δy = log[a(x + δx) + b] − log(ax + b)
= log
a(x + δx) + b
ax + b
= log
(ax + b) + aδx
ax + b
aδx 

= log 1 +
ax
+ b 

δy 1
aδx 

= log 1 +

δx δx
 ax + b 
∴
=
a
ax + b
aδ x 

⋅
log 1 +

ax + b aδx
 ax + b 
a
a δx 

=
log 1 +

ax + b
 ax + b 
a 

Multiply and divide by ax+b 


ax + b
aδx
Taking limits on both sides as δx → 0
∴
δy
a
a δx 

lim
=
lim log 1 +
δx →0 δx ax + b δx → 0
ax
+ b 

or
dy
a
=
loge
dx ax + b
or,
dy
a
=
dx ax + b
MATHEMATICS
ax + b
aδ x
1


x


(1 + x ) =e 
Q xlim
→0

283
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
Working rule :
d
1 d
log(ax + b) =
(ax + b)
dx
ax + b dx
=
Notes
Example 23.6
Find the derivative of each of the functions given below :
(iii) y = ( logx)3
(ii) y = log x
(i) y = logx5
Solution : (i)
1
a
×a =
ax + b
ax +b
y = logx5 = 5 log x
dy
1 5
= 5⋅ =
dx
x x
∴
y = log x = logx 2 or
∴
dy 1 1 1
= ⋅ =
dx 2 x 2x
(iii)
y = ( logx)
∴
y = t3 ,
⇒
dy
dt 1
= 3t 2 and
=
dt
dx x
3
We know that,
when
t = logx
dy dy dt
1
=
⋅
=3t 2 ⋅
dx dt dx
x
∴
dy
1
= 3(logx)2 ⋅
dx
x
∴
dy 3
= (logx)2
dx x
Example 23.7 Find,
(i)
1
y = logx
2
1
(ii)
dy
if
dx
y = x 3 logx
(ii)
y = e x logx
Solution :
(i)
∴
y = x 3 logx
dy
d
d
= logx (x 3 ) + x 3 (logx)
dx
dx
dx
= 3x 2 logx + x 3 ⋅
[Using Product rule]
1
x
= x 2 ( 3logx + 1)
284
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
(ii)
y = e x logx
∴
dy
d
d
= ex
(logx) + logx ⋅ e x
dx
dx
dx
= ex ⋅
MODULE - V
Calculus
1
+ ex ⋅ logx
x
Notes
1

= e  + logx
x


x
Example 23.8 Find the derivative of each of the following functions :
(i)
log tan x
Solution : (i) Let
(ii)
log [cos (log x )]
y = log tan x
dy
1
d
=
⋅ (tanx)
dx t a n x dx
∴
=
1
⋅ sec2 x
tanx
=
cosx
1
⋅ 2
s i n x cos x
= cosec x.sec x
(ii) Let
y = log [cos (log x)]
∴
dy
1
d
=
⋅ [cos(logx)]
dx cos(logx) dx
d


⋅−
 sinlogx dx ( logx) 
=
1
cos (logx )
=
− sin(logx) 1
⋅
cos(logx) x
1
= − tan(logx)
x
Example 23.9 Find
Solution :
∴
MATHEMATICS
dy
, if y = log(sec x + tan x)
dx
y = log (sec x + tax x )
dy
1
d
=
⋅
( s e c x + t a n x)
dx secx + t a n x dx
=
1
⋅ sec x t a n x + sec2 x 

secx + t a n x 
=
1
⋅secx [sec x+ t a n x]
secx + t a n x
285
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
secx(tanx + secx)
secx + tanx
= sec x
=
Example 23.10 Find
dy
, if
dx
Notes
1
y=
(4x 2 − 1)(1 +x 2) 2
3
x 3 (x − 7) 4
Solution : Although, you can find the derivative directly using quotient rule (and product rule)
but if you take logarithm on both sides, the product changes to addition and division changes to
subtraction. This simplifies the process:
y=
(4x − 1)(1 +x
2
x
3
1
2 2
)
3
(x − 7) 4
Taking logarithm on both sides, we get

 4x 2 − 1 1+ x 2
logy = log 
3

x 3 (x − 7 ) 4


(
∴
or
(
)(
)
)
(
1
2






)
1
3
logy = log 4x 2 − 1 + log 1 + x 2 − 3logx − log(x − 7)
2
4
[ Using log properties]
Now, taking derivative on both sides, we get
d
1
1
3 3  1 
(logy) = 2
⋅ 8x +
⋅ 2x − − ⋅ 

2
dx
x 4 x − 7 
4x − 1
2(1 + x )
⇒
∴
1 dy
8x
x
3
3
⋅ =
+
− −
2
2
y dx 4x −1 1 + x
x 4 (x − 7)
dy
 8x
x
3
3 
= y 2
+
− −
2
dx
x 4(x − 7) 
 4x − 1 1 + x
=
286
(4x 2 − 1) 1 + x 2  8x
x
3
3 
+
− −
 2
3
2
x 4(x − 7) 
 4x − 1 1 + x
3
x (x − 7) 4
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
CHECK YOUR PROGRESS 23.2
1.
Find the derivative of each the functions given below:
(a) f (x) = 5 sin x- 2 log x
2.
Find
(b) f (x) = log cos x
dy
, if
dx
Notes
2
x
(b) y = e
logx
2
(a) y = ex logx
3.
Find the derivative of each of the following functions :
π x
(b) y = log tan  + 
(a) y = log ( sin log x )
4
a + btanx
 a − b t a n x 

(c) y = log 
4.
Find
2
(d) y = log (log x )
dy
, if
dx
(a) y = (1
1
+x) 2 (2
2
−x) 3 (x 2
1
+5) 7 (x
3
−
+
9) 2
(b) y =
3
x(1 − 2x) 2
5
(3 + 4x) 4 (3
1
2 4
−7x )
23.3 DERIVATIVE OF LOGARITHMIC FUNCTION (CONTINUED)
We know that derivative of the function x n w.r.t. x is n x n −1 , where n is a constant. This rule is
not applicable, when exponent is a variable. In such cases we take logarithm of the function and
then find its derivative.
Therefore, this process is useful, when the given function is of the type [f ( x )]g( x ) . For example,
a x , x x etc.
Note : Here f(x) may be constant.
Derivative of ax w.r.t. x
y = ax ,
Let
a>0
Taking log on both sides, we get
logy = loga x = xloga
∴
d
d
( logy ) = (xloga)
dx
dx
or
MATHEMATICS
[ logm n = nlogm]
or
1 dy
d ( )
⋅ = loga×
x
y dx
dx
dy
= yloga
dx
287
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
= a x loga
d x
a = a x loga ,
dx
Thus,
a>0
Example 23.11 Find the derivative of each of the following functions :
(ii) y = x sinx
(i) y = x x
Notes
Solution : (i) y = x x
Taking logrithms on both sides, we get
logy = xlogx
Taking derivative on both sides, we get
1 dy
d
d
⋅
= logx (x) + x (logx)
y dx
dx
dx
[Using product rule]
1 dy
1
⋅
= 1 ⋅ logx + x ⋅
y dx
x
= log x + 1
∴
dy
= y[logx + 1]
dx
Thus,
dy
= x x (logx + 1)
dx
(ii)
y = x sinx
Taking logarithm on both sides, we get
log y = sin x log x
∴
1 dy d
⋅ = (sinxlogx)
y dx dx
or
1 dy
1
⋅
= cosx.logx + sinx ⋅
y dx
x
or
dy
sinx 

= y cosxlogx +
dx
x 

Thus,
dy
sinx 

= x sinx cosxlogx +
dx
x 

Example 23.12 Find the derivative, if
(
y = (logx ) + sin − 1 x
x
)
sinx
Solution : Here taking logarithm on both sides will not help us as we cannot put
288
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
(logx)x + (sin −1 x)s i n x in simpler form. So we put
u = (logx) x
and
v = (sin − 1 x)s i n x
y= u+ v
Then,
dy du dv
=
+
dx dx dx
∴
......(i)
Notes
u = (logx) x
Now
Taking log on both sides, we have
logu = log(logx) x
logu = xlog(logx)
∴
Q log mn = n l o g m


Now, finding the derivative on both sides, we get
1 du
1 1
⋅
= 1 ⋅ log(logx) + x
⋅
u dx
logx x
Thus,
du

1 
= u log(logx) +

dx
logx


Thus,
du

1 
= (logx)x log(logx) +
dx
logx

.....(ii)
v = (sin − 1 x)s i n x
Also,
logv = sinxlog(sin −1 x)
∴
Taking derivative on both sides, we have
d
d
(logv) = [sinxlog(sin −1 x)]
dx
dx
1 dv
1
1
= sinx ⋅ − 1 ⋅
+ cosx ⋅ log (sin − 1 x )
2
v dx
sin x 1 − x
or,
dv
sinx


= v
+ cosx ⋅ logsin − 1 x 
−1
2
dx
 sin x 1− x

sinx 
= ( sin −1 x )
sinx

+cosxlog ( sin −1 x ) 
 −1
2
 sin x 1− x

....(iii)
From (i), (ii) and (iii), we have
dy
1  ( −1 )sinx 
sinx
x

= ( logx ) log ( logx ) +
+ sin x
+ cosxlogsin −1 x 


−
1
2
dx
logx 

 sin x 1− x

Example 23.13 If x y = e x −y , prove that
dy
logx
=
dx (1 + l o g x)2
MATHEMATICS
289
Differentiation of Exponential and Logarithmic Functions
MODULE - V Solution : It is given that
Calculus
x y = e x− y
...(i)
Taking logarithm on both sides, we get
ylogx = (x − y)loge
= (x − y)
or
y(1 + logx) = x
Notes
y=
or
[Q
l o g e = 1]
x
1 + logx
(ii)
Taking derivative with respect to x on both sides of (ii), we get
dy
=
dx
=
Example 23.14 Find ,
1
(1 + logx ) ⋅−
 1
x 
x
(1 + l o g x)2
1 + logx −1
(1 + l o g x )
2
=
logx
(1 +l o g x )2
dy
if
dx
e x l o g y = sin − 1 x +sin − 1 y
Solution : We are given that
e x l o g y = sin − 1 x +sin − 1 y
Taking derivative with respect to x of both sides, we get
1
1
dy
x  1 dy 
x
e 
+ e l o g y=
+

ydx 
1 − x2
1 − y2 dx
or
ex
1  dy
1
=
−e x l o g y
 −

2
2
y
dx

1 − y 
1− x
or
2 
x
2

dy y 1− y 1 − e 1 − x l o g y
=
dx
e x 1 − y 2 − y 1 − x 2


Example 23.15 Find
dy
( cosx )......∞
, if y = ( c o s x )( c o s x)
dx
Solution : We are given that
cosx )......∞
cosx )(
y
y = (cosx )(
=(c o s x)
Taking logarithm on both sides, we get
290
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
logy = y logcosx
Differentiating (i) w.r.t.x, we get
1 dy
1
dy
=y ⋅
(− s i n x )+ log (c o s x)⋅
y dx
cos x
dx
or
1
 dy
 y − log (c o s x ) dx = −y t a n x


or
dy
1 − ylog (c o s x )
= −y 2 t a n x
dx
Notes
dy
− y2 t a n x
=
dx 1 − ylog (c o s x)
or
CHECK YOUR PROGRESS 23.3
1.
Find the derivative with respect to x of each the following functions :
(a) y = 5 x
2.
3.
4.
(b) y = 3x + 4 x
(c) y = sin(5 x )
(a) y = x 2 x
(b) y = (cosx)logx
(c) y = (logx)s i n x
(d) y = (tanx)x
(e) y = (1 +x 2) x
Find
dy
, if
dx
2
(f) y = x (x
2
+sinx)
Find the derivative of each of the functions given below :
−1
(a) y = (tanx)cotx + (cotx)x
(b) y = x logx + (sinx)sin
(c) y = x tanx + (sinx)cosx
(d) y = (x) x + (logx)logx
If
x
2
sinx......
) ∞
s i n x )(
, show that
y = (s i n x)(
dy
y 2 cot x
=
dx 1 − ylog ( s i n x)
5.
If y = logx + logx + logx ......
+
∞, show that
dy
1
=
dx x (2x − 1)
MATHEMATICS
291
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
Notes
23.4 SECOND ORDER DERIVATIVES
In the previous lesson we found the derivatives of second order of trigonometric and inverse
trigonometric functions by using the formulae for the derivatives of trigonometric and inverse
trigonometric functions, various laws of derivatives, including chain rule, and power rule discussed
earlier in lesson 21. In a similar manner, we will discuss second order derivative of exponential
and logarithmic functions :
Example 23.16 Find the second order derivative of each of the following :
(i) e x
(iii) x x
(ii) cos (logx)
Solution : (i) Let y = e x
Taking derivative w.r.t. x on both sides, we get
Taking derivative w.r.t. x on both sides, we get
∴
d2 y
dx
2
dy
= ex
dx
d2 y
dx
2
=
d x
(e ) = ex
dx
= ex
(ii) Let
y = cos (log x)
Taking derivative w.r.t x on both sides, we get
dy
1 − sin (l o g x)
= −sin (l o g x) ⋅ =
dx
x
x
Taking derivative w.r.t. x on both sides, we get
d2 y
dx 2
d  sin (l o g x) 
−

dx 
x

1
x ⋅ cos ( logx⋅) − sin (l o g x)
x
= −
2
x
or
∴
=
d 2y
dx 2
=
sin (l o g x ) − cos (l o g x)
x2
(iii) Let y = x x
Taking logarithm on both sides, we get
log y = x log x
Taking derivative w.r.t. x of both sides, we get
....(i)
1 dy
1
⋅
= x ⋅ + logx = 1 + logx
y dx
x
or
dy
= y(1 + logx)
dx
....(ii)
Taking derivative w.r.t. x on both sides we get
292
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
d2 y
d
= [ y(1 + logx) ]
2
dx
dx
1
x
= y ⋅ + (1+ logx)
dy
dx
=
y
+(1 +logx)y(1 logx)
+
x
=
y
+(1 +logx) 2 y
x
...(iii)
Notes
(Using (ii))
1

= y  + (1 + logx)2 
x


d 2y
1
2
= x x  +(1 +l o g x) 
2
x
dx


∴
Example 23.17 If y = ea c o s
(
1− x2
−1 x
, show that
− a 2y = 0 .
) dx 2 − x dy
dx
d2 y
Solution : We have, y = ea c o s
−1 x
.....(i)
−1
dy
−a
= ea c o s x ⋅
dx
1− x2
∴
= −
ay
Using (i)
1 − x2
2
or
a 2 y2
 dy
=
 dx
  1−x2
2
∴
(
)
 dy
2
2 2
 dx 1 − x − a y = 0
 
.....(ii)
Taking derivative of both sides of (ii), we get
 dy 
 dx 
 
or
MATHEMATICS
(
2
1− x2
( − 2x ) + 2 (1− x 2 )×
dy d 2 y
dy
⋅ 2 − a 2 ⋅ 2y ⋅
=0
dx dx
dx
− a 2y = 0
) dx 2 − x dy
dx
d2 y
[Dividing through out by2 ⋅
dy
]
dx
293
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
CHECK YOUR PROGRESS 23.4
1.
Find the second order derivative of each of the following :
5x
(b) tan ( e )
(a) x 4 e5x
Notes
2.
If
logx
x
y = acos ( logx ) + bsin ( l o g x) , show that
x2
d2 y
dx
3.
(c)
If y = e tan
(
2
−1 x
1+ x2
+x
dy
+y =0
dx
, prove that
) dx 2 +(2x
d2 y
−1)
dy
=0
dx
LET US SUM UP
d x
(e ) = e x
dx
l
(i)
(ii)
l
If µ is a derivable function of x, then
d x
(a ) = ax loga ; a > 0
dx
(i)
d µ
dµ
(e ) = eµ ⋅
dx
dx
(iii)
d ax + b
(e
) = eax + b ⋅ a = aeax + b
dx
l
(i)
d
1
(logx) =
(ii)
dx
x
l
(iii)
d
1
a
log(ax + b) =
⋅a =
dx
ax + b
ax + b
(ii)
d µ
dµ
(a ) = aµ ⋅ loga ⋅
;a > 0
dx
dx
d
1 dµ
(log µ ) = ⋅
, if µ is a derivable function of x.
dx
µ dx
SUPPORTIVE WEB SITES
l
l
294
http://www.wikipedia.org
http://mathworld.wolfram.com
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
TERMINAL EXERCISE
1.
Find the derivative of each of the following functions :
xx )
(
(b) x
x x
2.
(a) (x )
dy
Find
, if
dx
cos
(b) y = (sinx)
(a) y = a x l o g s i n x
1
(c) y = 1 + 
 x
3.
Find the derivative of each of the functions given below :
(b) f(x) = (sin −1 x) 2 ⋅ x s i n x ⋅ e 2x
2
Find the derivative of each of the following functions :
(a) y = (tanx) logx + (cosx)s i n x
5.
Find
Find
If y =
(3x + 5) 2
If y = a x ⋅ xa
(b)
If y =
(b)
y = 7x
e x + e− x
(ex − e− x )
2 +2x
Find the derivative of each of the following functions :
(a) y = x e
2 2x
8.
x4 x + 6
dy
, if
dx
(a)
7.
(b) y = x t a n x + (sinx)c o s x
dy
, if
dx
(a)
6.
−1 x
3

x
−
4
 4
 x
(d) y = log e 

x +4  


x2
(a) f(x) = cosxlog(x)e x x x
4.
Notes
x
If y = x
cos3x
x x.......∞
MATHEMATICS
(b) y =
, prove that
x
2x c o t x
x
dy
y2
=
dx 1 − y l o g x
295
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
ANSWERS
CHECK YOUR PROGRESS 23.1
7
1.
(a) 5e5x
2.
(a) ex
Notes
(b) 7e 7x + 4
1
3
(c)
2e
2x
2
(b) sec x + 2 c o s x− 3 s i n x −
(d) −
7 −2 x
e
2
2
(e) 2 ( x +1 ) e x + 2x
1 x
e
2
(c) 5cosx − 2ex (d) ex − e−x
3.
(a)
e x +1
2 x+1
(c) e x s i n
2x
(b) e
cotx
 − cosec 2 x 


 2 c o t x 
[sinx + 2xcosx]sinx
2
(d) e xsec x [sec 2 x + 2xsec 2 x t a n x]
4.
5.
(b) 2e 2x sinx(sinx + cosx)
(a) xex
(a)
2x 2 − x + 2
(x 2 + 1)3 / 2
e 2x
(b)
e 2 x [(2x − 1)cotx − xcosec 2 x]
x2
CHECK YOUR PROGRESS 23.2
1.
(a) 5cosx − 2
2.
(a) e
3.
(a)
(c)
4.
(b) − tanx
x
x2

 2xlogx +

1
x 
cot(logx)
x
2
2
1
2
a − b tan x
(d) xlogx
 1
2
2x
3 
1
2
1
3
−

(a) (1 + x) 2 (2 −x) 3 (x 2 +5) 7 (x +9) − 2 ×  2(1 + x) − 3(2 − x) +
2
7(x − 5) 2(x + 9) 

3
x(1 − 2x) 2
(b)
296
x(logx)2
.ex
(b) secx
2absec 2 x
2
(b)
2x 2 logx − 1
5
(3 + 4x) 4 (3
1
−7x 2 ) 4
1

3
5
7x
−
+
 −

2
 2x 1 − 2x 3 + 4x 2(3 − 7x ) 
MATHEMATICS
Differentiation of Exponential and Logarithmic Functions
CHECK YOUR PROGRESS 23.3
1.
2.
(b) 3x log3 + 4 x log4
(a) 5 x log5
(c) cos5 x 5x log5
 logcosx

− tanxlogx 
x


logx
(b) (cosx)

(a) 2x 2x (1 + logx)
sinx 
sinx 
 cosxlog ( l o g x ) + x l o g x 


(c) ( l o g x )
MODULE - V
Calculus
Notes
x
x 

(d) ( t a n x ) l o g t a n x +
s i n x c o s x

2

x3 

1 + x 2 
(e) (1 + x) x  2xlog(1 + x 2 )+ 2

(f) x (x
3.
2 + sinx)
 x 2 + sinx

+ (2x + cosx)logx

x


(a) cosec 2 x (1 − logtanx ) ( tanx )cotx + ( logcotx −xcosec 2 xtanx ) ( cotx )x
(b) 2x ( logx −1) logx + ( sinx )sin
−1 x
logsinx 

−1
cotxsin x +

1 − x2 


(c) x tanx 
tanx

cosx
+ sec 2 xlogx +( sinx )
[cosxcotx −sinxlogsinx]
 x

x
logx 1 + log ( logx ) 
(d) ( x ) ⋅ x (1+ 2logx )+ ( logx)


2

x

CHECK YOUR PROGRESS 23.4
1.
(
5x
25x 4 + 40x 3 +12x 2
(a) e
(c)
)
( ){
5x
2 5x
1 + 2e5x t a n e5x
(b) 25e sec e
}
2logx −3
x3
TERMINAL EXERCISEs to Terminal uestions
(a) (x x ) x [x + 2xlogx]
2.
(a) a xlogsinx[logsinx + xcotx]loga
cos
(b) (s i n x )
MATHEMATICS
−1 x
x
(b) x (x) [x x−1 + logx(logx + 1)x x ]
1.
 −1
l o g s i n x
cos x c o t x −


1 − x 2 
297
Differentiation of Exponential and Logarithmic Functions
MODULE - V
Calculus
1
(c) 1 + 
 x
3.
Notes
x2



1
1 

2xlog  x +  −

x  1+ 1 



x 
(a) cosxlog(x)e x ⋅ x x − tanx+


2
2
−1
 1 − x sin x
4.
3
3
−
4(x − 4) 4(x + 4)
1

+ 2x + 1 + logx 
xlogx

2
(b) (sin − 1 x)2 .x sinx e 2 x 
(d) 1 +
+ cosxlogx +

sinx
+ 2
x

(a) (tanx) logx 2 c o s e c 2 x l o g x + logtanx 
x


1
+ (cosx)sinx [− sinxtanx + cosxlog(cosx)]
(b) x tanx 
tanx

+ sec2 xlogx  + (sinx)cosx [cotxcosx − sinxlogsinx)]
 x

5.
x4 x + 6  4
1
6 
+
−
(a)
2  x 2(x + 6) (3x + 5) 
(3x + 5) 

(b)
6.
(a) a x .x a −1[a + xlog e a]
(b) 7 x
7.
2 2x
(a) x e cos3x  x + 2 − 3tan3x 
2

(b)
298
−4e 2x
(e 2x − 1)2
2
+ 2x
(2x + 2)log e 7


2x cotx 
1 
log2 − 2cosec2x −

2x 
x 
MATHEMATICS