math

1. 1
Differentiation
Introduction:The derivative is a mathematical operator, which measures the rate of change of a
quantity relative to another quantity. The process of finding a derivative is called differentiation.
There are many phenomena related changing quantities such as speed of a particle, inflation of
currency, intensity of an earthquake and voltage of an electrical signal etc. in the world. In this
chapter we will discuss about various techniques of derivative.
Outcomes: After successful completion of the chapter, the students will be able to:
1. determine the speed, velocity and acceleration of a particle with respect to time.
2. calculate the rate at which the number of bacteria , the population changes with time.
3. measurethe rate at which the length of a metal rod changes with temperature.
4. find out the rate at which production cost changes with the quantity of a product .
Derivatives of elementary functions:
1.   0,
d
c
dx
 where c is a constant. 2.   1.
d
x
dx

3.   1
.
n n
d
x nx
dx

 4.   1
.
2
d
x
dx x

5.   .
x x
d
e e
dx
 6.   ln .
x x
d
a a a
dx

7.  
1
ln .
d
x
dx x
 8.  
sin cos .
d
x x
dx

9.  
cos sin .
d
x x
dx
  10.   2
tan sec .
d
x x
dx

11.  
sec sec tan .
d
x x x
dx
 12.   2
cot cos .
d
x ec x
dx
 
13.  
cos cos cot .
d
ecx ecx x
dx
  14.  
1
2
1
sin .
1
d
x
dx x



15.  
1
2
1
cos .
1
d
x
dx x
 


16.  
1
2
1
tan .
1
d
x
dx x



17.  
1
2
1
cot .
1
d
x
dx x
 


18.  
1
2
1
sec .
1
d
x
dx x x



19.  
1
2
1
cos .
1
d
ec x
dx x x
 


20.   .
d dv du
uv u v
dx dx dx
 
21.    
ln .
v v
d d
u u v u
dx dx
 22. 2
du dv
v u
d u dx dx
dx v v

 

 
 
where u and v are functions of x.

2. 2
 Find the differential coefficient (
dy
dx
)of the following functions with respect to x.
 
     
 
6
6
7
7
7
7
3 2 1
: , 3 2 1
2.
3
21 2 0
21 2 .
,
3 2 1
2 1
y x x
Given that y x x
Differentiating with res
dy d
dx dx
d d d
pect to x we get
x
dx dx dx
x
x A
x
ns
x
x
  
  



  
 

 
Sol
 
 
 
8
8
8 1
8
7
8
5
: , 5
1.
5
,
5
5 8
40 .
y x
Given that y x
Differentiating with
dy d
x
dx dx
d
x
dx
respect to x we ge
x
x ns
t
A



 



Sol
 
   
 
 
3.
4 sin cos
4cos sin
4c
4sin cos
: , 4sin cos
,
4sin cos
os sin .
y x x
Given that y x x
Differentiating with respect to x we g
dy d
dx dx
d d
x x
dx dx
x x
x x An
e
x
s
t
x

 
 
 
 


 
Sol
 
   
   
   
2 2
2 2
2 2
2 2
2
2
2
sec tan
: , sec tan
,
sec tan
sec tan
4.
2se sec tan
sec t
c 2tan
2sec an sec
sec
2tan
2sec tan 2 tan
0
y x x
Given that y x x
Differentiating with respect to x we get
x x
x x
x x
x
dy d
dx dx
d d
dx dx
d d
x x
x x
x
dx dx
x x
x x x
Ans

 
 
 
 

 
 

Sol
 
.
 
 
 
 
 
 
 
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2 2 2
2 2 2
ln
: , ln
5.
1
.
1
. 1
1 1
. 1 .
2
1 2
. 1
,
l
2
n
y x x a
Given that y x x a
Differentiating with respect to x we get
x x a
dy d
dx d
x x a
x x a
x a
x x a
x a
x x a x a
x x a
x
d
dx
d
dx
d
dx
x
x


 
 
 
 
 
 
 

  
  
 
 
 

 

  



 

Sol
 
2
2 2 2 2
2 2
2 2 2 2
2 2
1
. 1
1
.
1
.
a
x x a x a
x a
x x a x a
x
x
x
A s
a
n
 
 
 
 
 
 
 


  

 



 
 
 
 
 
 
 
 
2
ln sec tan
: , ln sec tan
,
ln sec tan
sec tan
sec tan
sec tan sec
sec tan
sec tan sec
sec t
6.
1
.
a
ec
.
n
s
dy d
dx dx
d
y x x
Given that y x x
Differentiating with respect to x we get
x x
x x
x x
x x x
x x
x x x
dx
x
An
x x
s
 
 












Sol

3. 3
 
 
 
 
 
cot
cot
cot
cot
cot
cot
2
cot 2
8.
. cot
1
. . cot
2 cot
. cos
2 cot
cos
2 c
: ,
,
ot
.
x
x
x
x
x
x
x
e
e
dy d
e
dx dx
d
e
y
Given that y
Differentiating with respect
x
dx
d
e x
dx
x
e
ec x
x
e
to x we ge
ec x
t
x
Ans



 


 
Sol
 
 
 
 
   
 
 
 
 
 
 
 
   
2
2
2
2 2
2 2
2
2 2 2
2
2
2
2
10.
sec .
1
sec . .
1
se
tan lnsin
: , tan lnsin
,
tan lnsin
lns
c .c
in ln sin
lnsin sin
sin
lnsin . .
sin
os
cot
x
x
x
x x
x x
x
x x x
x
x
y e
Given that y e
Differentiating with respect to x we get
e
dy d
e e
e e
dx dx
e
e e e
e
d
dx
d
dx
d
dx
e







Sol
     
   
 
2 2 2
2 2 2
2
2
2
sec
2 cot sec
.
lnsin .
lnsin
x x
x x x
e e x
e
d
d
x e
x
e
Ans

 
 
 
 
3
3
3
3
3
2
3
2
3
2 5
: , 2 5
,
2 5
1
2 5
2 2 5
1
3 2 0
2 2 5
3 2
2 2 5
9.
.
.
.
y x x
Given that y x x
Differentiating with respect to x we get
dy d
dx dx
d
dx
Ans
x x
x x
x x
x
x x
x
x x
  
  
 
 
 
 
 




 

Sol
 
 
 
 
 
 
1
1
1
1
1
1
1
1
1
1
1 cot
1 cot
1 cot
cot
2cot
cot
1
2cot
cot
2
2cot
cot
cos
: , cos
,
cos
1
cot
1
11.
1
1
1
1
1
.
.
x
x
x
x
x
x
x
x
x
x
y e
Given that y e
Differentiating with respect to x we g
dy d
dx dx
d
dx
d
d
et
e
e
e
e
e
x
x
e
x
e
e


















 

 

 


 

 



Sol
 
 
1
2 2cot
1
.
x
s
x e
An


 
   
 
 
1
1
1
1
1
1
sin 1
si
2
n 1
sin 1
sin 1
sin 1
sin
2 2
12.
1
1
1
1
.
tan
: , tan
,
tan
tan
. sin
1
x
x
x
x
x
x
dy d
dx dx
d
dx
y e x
Given that y e x
Differentiating with respect to x we get
e x
d
e x
dx
d
e x
dx
e
x
x
An
x
s











 
 











Sol
 
 
 
 
 
2
2
2
2
2
2
2
7.
.
.
: ,
,
2 0
2
ax bx c
ax bx c
ax bx c
ax bx c
ax bx c
ax bx c
y e
Given that y e
Differentiating with respect to x we ge
dy d
dx dx
d
dx
An
t
e
e ax bx c
e ax b
ax b e
s
 
 
 
 
 
 






 
 

Sol

4. 4
 
   
 
 
2 1
2 1
2 1
2 1 1 2
2 1
2
2
1
2
cot
: , cot
,
cot
cot cot
1
cot 2
1
2 cot
13
1
.
.
dy
dx
y x x
Given that y x x
Differentiating with respect to x we get
d
x x
dx
d d
x x x x
dx dx
x x x
x
x
x
s
x
n
x
A



 






 

 






 

Sol
 
   
 
 
3
3
3
3 3
3 2
2 2
ln
: , ln
,
ln
ln ln
1
. ln 2
2 n
4.
l
1
.
y x x
Given that y x x
Differentiating with respect to x w
dy
e get
d
x x
dx
d d
x x x x
dx dx
x x x
x
dx
Ans
x x x









Sol
 
   
   
 
 
sin
: , sin
,
sin
sin sin
cos sin
cos sin
cos sin n
1 .
s
.
i
5 x
x
x
x x
x x x
x x x
x x x
y xe x
Given that y xe x
Differentiating with respect to x we get
d
xe x
dx
d d
xe x x xe
dx dx
d d
xe x x x e e x
dx dx
xe x x xe e
xe x xe x e x
dy
dx
Ans



 
 
 
 
 







Sol
 
   
   
 
cos
: , cos
,
cos
cos cos
sin cos
cos sin
16.
.
ax
ax
ax
ax ax
ax ax
ax ax
dy
dx
y e bx
Given that y e bx
Differentiating with respect to x we get
d
e bx
dx
d d
e bx bx e
dx dx
e b bx bx ae
ae bx
A
b
ns
e bx






 
 
Sol
 
 
 
 
 
   
   
2
2
2
2
2
2 1 1
2 1 1
2 1 1
2 1 1
2 1 1
2 2
2 1
1
2
1 sin
: , 1 sin
,
1 sin
1 sin
1 1
1 . sin . 2 . .2
1 2 1
1
2 i
1 .
s n
1
7 x
x
x
x
x
y x x e
Given that y x x e
Differentiating with respect to x we get
d
x x e
d
dy
dx x
d d
x x e
dx dx
x x x e x
x x
x xe
x x
x
 
 
 
 
 




  
  
 
 
  
 



 
Sol
 
2
2
.
1
x
x
Ans


 
 
 
 
 
     
     
   
sin
sin
sin
sin sin
sin sin
sin sin
sin
: , sin
,
sin
sin sin
.cos . sin . .cos
.cos . l
1
n sin . .co
8.
s
x x
x x
x x
x x x x
x x x x x
x x x x x
y e a
Given that y e a
Differentiating with respect to x we get
d
e a
dx
d d
e a a e
dx dx
d
e a a a e x
dx
e a a a a e x
dy
dx
An









Sol
 
.
s

5. 5
1
2
1
2
1
1
2
1
2
1
1
1
tan
1
: , tan
1
, sin sin
sin
, tan
1 sin
sin
tan
cos
sin
tan
cos
tan .tan
sin
21.
x
y
x
x
Given that y
x
put x x
Now y
x
Differentiating with respect to x we ge
 
















 
  

 
 
  

 
  
 
  

 
 
  
 
 
  
 



Sol
 
 
1
2
,
sin
.
1
1
t
dy
dx
Ans
d
x
dx
x




     
 
 
 
 
2
2
2
2
2
2
2
cos
1 sin
cos
: ,
1 sin
,
cos
1 sin
1 sin cos 1 sin
1 sin
cos
1 sin
c
19.
cos
sin sin
sin sin
sin 1
os
1 sin
1
dy d
dx
x
y
x
x
Given that y
x
Differentiat
dx
d d
x
dx dx
x
ing with respect to x we get
x
x
x x x
x
x
x
x
x
x
x x
x









 
  

  

 


 
 



Sol
 
 
2
1
1 s
sin
in
.
x
x
Ans
 


       
 
 
2
cos sin
cos sin
cos sin
: ,
cos sin
,
cos sin
cos sin
cos sin cos sin cos sin cos sin
cos sin
cos sin sin
20.
dy d
d
x x
y
x x
x x
Given that y
x x
Differentiating with respect to x we get
x x
x x
x x x
x dx
d d
dx dx
x x x x x
x x
x x x







 
 






   


  
Sol
    
   
   
 
2
2 2
2
cos cos sin sin cos
cos sin 2sin cos
cos sin cos sin
1 sin2
1 sin2 1 sin2
1 sin2
1 sin
2
2
.
x x x x x
x x x x
x x x x
x
x x
x
x
Ans

 
  
 
 


 

 
 


 
 
2
1
2
2
1
2
1
2
1
2
1
1
1
2
1
cos
1
1
: , cos
1
, tan tan
1 tan
, cos
1 tan
cos .cos2
2
2tan
,
2
22
tan
1
.
2
.
x
y
x
x
Given that y
x
put x x
Now y
x
Differentiating with respect to x we get
x
x
dy d
dx dx
Ans
 











 

  

 
 

  

 
  
 

  

 






Sol

6. 6
 
 
 
2
1
2
1
1
2
1
2
1
1
1
1 1
tan
1 1
: , tan
, tan tan
1 tan 1
, tan
tan
sec 1
tan
tan
sec 1
tan
tan
1 cos cos
tan .
cos
2
s n
3
t
.
i
x
y
x
x
Given that y
x
put x x
Now y
 






 
 







 
 
 

 
 
 
 
 
 

 
 
 
  
 
 
 

 
 
 
 

  
 
 

 
  
 

 
  
 

Sol
 
 
 
1
1
2
1
1
1
1
1
2
1 cos
an
sin
1 cos
tan
sin
2sin
2
tan
2sin cos
2 2
sin
2
tan
cos
2
tan .tan
2
2
1 tan
2
,
1 tan
2
1
2 1
.
x
Differentiating with respect to
dy
dx
Ans
x we get
d
x
dx
x





 












 
 
 

 
  
 
 
 

 
 
 
 

 
 







7. 7
Homework:-Find
dy
dx
of the following functions:
1.  
ln
y x a x b
    Ans:
  
1
2 x a x b
 
2.    
cos ln ln tan
y x x
  Ans:
 
sin ln
2cos 2
x
ec x
x

3. 𝑦 = 𝑠𝑖𝑛−1
(𝑒𝑡𝑎𝑛−1𝑥
)
4. sin
ax m
y e rx
 Ans:  
sin cot
ax m
e rx a mr rx

5.
2
1 2
sin x
y x xe

  Ans:  
2
2
4
2
2 1
1
x
x
x e
x
 

6. 1
2
tan
1
x
y
x
  
  

 
Ans:
2
1
1 x

7.
1 4
tan
1 4
x
y
x

 
  
 

 
Ans:
 
2
1 4
x x

8.
1 cos
tan
1 sin
x
y
x
  
  

 
Ans:
1
2

Logarithmic differentiation: If we have functions that are composed of products, quotients
and powers, to differentiate such functions it would be convenient first to take logarithm of the function
and then differentiate. Such a technique is called the logarithmic differentiation.
 
 
 
 
   
 
   
     
   
 
ln
ln
ln
ln
ln
ln
ln
: ,
1. sin
sin
sin
sin ln ln sin
sin ln . sin sin .
sin ln . sin .
sin co
,
ln ln ln
1 1
.cos ln
sin
t
x
x
x
x
x
x
x
Given that y
Differentiating with respect to x we get
d
dx
d
dx
y x
x
dy
x
dx
x x x
x x x
d d
x
dx dx
x
x x
x
x x x
x






 
  
 
 
  
 

Sol
 
 
in
ln
.
ln s
x
x
x x
Ans
 





 
 
 
     
   
 
2
2
2
2
2
2
2
1
1
1
1 2
1 2 2
1 2
1
2.
: ,
,
1 ln
ln 1 1 l
. .
ln . .
ln
n
1
0 1 2 1
2 1
x x
x x
x x
x x
x x
x
x
x
x
y x
Given that y x
Differentiating with respect to x we get
d
x
dx
d
x x x x
dx
d d
x x x x x x
dx d
dy
dx
x
x
x x
x
x x x x
x
x
 
 
 
 








 
  
 
 
  



 
    
    



Sol
 
 
2
1
.
x x
A s
x
n
 
 
 



 

8. 8
 
 
 
 
     
 
     
   
sin cos
1
sin cos
1
sin cos
1
sin cos
1 1
sin cos
1 1 1
3. tan
tan
tan
tan sin cos .ln tan
tan sin cos tan tan
: ,
,
ln ln .
x x
x x
x x
x x
x x
y x
x
dy
x
dx
x x x x
x
Given that y
Differentiating with respect to x we get
d
dx
d
dx
d d
dx d
x x x x
x







 

  



 



Sol
 
   
 
   
 
sin cos
1 2
1 1
sin cos
sin cos
tan tan . cos sin
ta
. ln
1
n
.
1
x x
x x
x x
x x x x
x
Ans
x

 

 

 
 
 

 
 
 




 
 
 
 
   
 
     
 
 
 
ln
ln
l
n
ln
ln
n
l
4. sin
sin
sin
sin
sin sin
sin
s
: ,
,
ln ln ln
ln
1 1
. ln ln . .co
s n
in s
i
x
x
x
x
x
x
x
x
x
x
x
x
Given that y
Differentiating with respect to x we get
d
dx
d d
x
dx dx
d d
x x x x
dx dx
x x x x x
x
y x x
x x
dy
x x
dx
x
x x
x
x
x x
 

 




 
 


 
 



 
 

Sol
   
 
ln ln
1 ln ln .co
sin
sin t .
x
x x
x x x x Ans
x
x
 
  
 
 
   
   
   
 
 
   
 
   
     
 
 
cos sin
cos sin
cos
cos sin
cos sin
cos sin
: ,
,
sin
sin cos ln sin
5. sin cos
sin cos
sin cos
c
si
os
cos co
n ln
sin cos
s
x x
x x
x x
x x
x x
x
Given that y
Differentiating with respect to x we get
d
dx
d d
x
dx dx
d d
x x x x
dx d
y x x
x x
dy
x x
dx
x
x x
x
x
 

 






Sol
       
       
sin
cos sin
1 1
. .cos sin ln sin sin . . sin cos ln cos
sin cos
sin cos .cot sin ln sin cos ln cos sin .ta
cos
cos n .
x
x x
x x x x x x x
x
x x
x x
x
x x x x x x x x Ans
 
   
 
 
   
  
 
 
 
  
 

9. 9
9. (𝑐𝑜𝑠𝑥)𝑦
= (𝑠𝑖𝑛𝑦)𝑥
Sol: Taking 𝑙𝑛 in both sides we get
𝑦𝑙𝑛𝑐𝑜𝑠𝑥 = 𝑥𝑙𝑛𝑠𝑖𝑛𝑦
Differentiating w.r.to 𝑥 we get
(𝑙𝑛𝑐𝑜𝑠𝑥)
𝑑𝑦
𝑑𝑥
+ 𝑦
1
𝑐𝑜𝑠𝑥
(−𝑠𝑖𝑛𝑥) = 1. 𝑙𝑛𝑠𝑖𝑛𝑦 + 𝑥
1
𝑠𝑖𝑛𝑦
𝑐𝑜𝑠𝑦
𝑑𝑦
𝑑𝑥
⟹ (𝑙𝑛𝑐𝑜𝑠𝑥 − 𝑥𝑐𝑜𝑡𝑦)
𝑑𝑦
𝑑𝑥
= (𝑙𝑛𝑠𝑖𝑛𝑦 + 𝑦𝑡𝑎𝑛𝑥)
or
𝑑𝑦
𝑑𝑥
=
(𝒍𝒏𝒔𝒊𝒏𝒚+𝒚𝒕𝒂𝒏𝒙)
(𝒍𝒏𝒄𝒐𝒔𝒙−𝒙𝒄𝒐𝒕𝒚)
Ans
Homework:-Find
dy
dx
of the following functions:
1.
1
sin x
y x

 Ans:
1
1
sin
2
sin ln
1
x x x
x
x x


 

 

 
2.  
1
cos
sin
x
y x

 Ans:  
1
cos 1
2
lnsin
sin cot cos
1
x x
x x x
x


 

 

 
 
   
 
1
1
1
1
1
1
cos
cos
cos
cos
cos 1
cos
1
2
:
6. sin ln
sin ln
sin ln
sin ln
sin
c
,
os cos ln
,
ln
cos ln
1
x
x
x
x
x
x
Given that y
Differentiating with respect to x we get
d
dx
d d
x
dx dx
y x x x
x x x
dy
x x x
dx
x x
x
x x
x
d
x x x
dx
x x
x
x x








 

 




 
 
 





Sol
sin
cos ln .
x
x x A
x
ns
 
 




 
 
   
   
   
 
 
   
 
   
     
 
 
tan ln
2
tan ln
2
tan ln
2
tan ln
2
tan ln
2 2
t
2
7. 1 2 sin
1 2 sin
1 2 sin
1 2 s
: ,
in
1 tan 1 2 sin ln 2 si
,
ln l n
n
1
x x
x x
x x
x x
x x
y x x
x x
dy
x x
dx
Given that y
Differentiating with respect to x we get
d
dx
d
dx
d d
x
dx d
d
x x
dx
x x x
x
x
x x
   
  
   
   
   

 
 
Sol
   
 
 
an ln
2
2
2 2 sin
1
ln
2 tan cos ln
sec ln
2 si
2 sin .
n
1
x x x x
x x
x x x
x
x
A s
x
n
x
 

 

 
    
 

   

10. 10
3.
x
x
y x
 Ans:
1
1 ln
x
x x
x x x
x
 
 
 
 
4.  
1 ln
cos
sin
x
x
y x x

  Ans:  
1
1
ln
cos
2
cos ln lnsin
sin ln cot
1
x
x x x x
x x x x
x x
x


   
  
   
 

 
5.    
cot tan
tan cot
x x
y x x
  Ans:        
cot tan
2 2
cos 1 lntan cot s lncot 1
x x
tanx ec x x x ec x x
  
6. 𝑥 + 𝑦 = tan(𝑥𝑦)
7. √
𝑦
𝑥
+ √
𝑥
𝑦
= 6
Parametric Equation: If in the equation of a curve  
y f x
 , x and y are expressed in terms of a
third variable known as parameter i.e,    
,
x t y t
 
  then the equations are called a parametric
equation.
   
 
 
 
 
 
2
1. sin , 1 cos
: ,
sin (1)
1 cos (2)
(1) (2) ,
1 cos
sin
,
sin
1 cos
2sin cos
2 2
2 cos
2
tan .
2
x a t t y a t
sol Given that
x a t t
and y a t
Differentiating and with respect to t we get
dx
a t
dt
dy
and a t
dt
dy
dy dt
Now
dx
dx
dt
a t
a t
t t
t
t
Ans
   
 
 
 






   
 
 
 
 
2. cos sin , sin cos
: ,
cos sin (1)
sin cos (2)
(1) (2) ,
sin cos sin
cos
cos sin cos
sin
,
x a t t t y a t t t
sol Given that
x a t t t
and y a t t t
Differentiating and with respect to t we get
dx
a t t t t
dt
at t
dy
and a t t t t
dt
at t
dy
dy dt
Now
d
dx
   
 
 
   

  


 
sin
cos
tan .
x
dt
at t
at t
t Ans



11. 11
 
1
1
1
1
2 sin
2
sin
2
2
2
2
sin
2
sin
2
sin
3. 1 ,
: ,
1 (1)
(2)
(1) (2) ,
1
1 . 2
2 1
1
1
1
1
1
.
1
1
,
t
t
t
t
x t t y e
sol Giventhat
x t t
and y e
Differentiating and with respect tot we get
dx
t
dt t
t
t
t t
t
dy
and e
dt t
e
t
dy
dy dt
Now
dx
dx
dt
e





   
  

  

 

 








 
1
1
2
2 2
sin
2
1
.
1 1
.
1
t
t
t
t t t
e
Ans
t t


  

 
1 1
2 2
1
2
1
2 1
1
1
1
2
1
2 2
4. tan sin .
1 1
2
: , tan
1
tan
2 tan
tan ;
1 tan tan
tan .tan 2
2
2 tan (1)
2
sin
1
2
sin
x x
Differentiate with respect to
x x
x
sol Let y
x
putting x
x
x
x
and z
x


 


 







   
   
 
   
 
  

 

 
 
  
 
  
   



 
  

 

 
2 1
1
1
2 2
2
2
tan
tan
;
1 tan tan
sin .sin 2
2
2 tan (2)
(1) (2) ,
2 2
1 1
,
2
1
2
1
1 .
putting x
x
x
Differentiating and with respect to x we get
dy dz
and
dx x dx x
dy
dy dx
Now
dz
dz
dx
x
x
Ans


 






 
 
 
 
  
   



 
 






12. 12
     
1
1
1
1
sin 1
sin
1
sin 1
1
sin
2
6. sin .
: , (1)
sin (2)
(1) (2) ,
sin ln ; ln
sin ln
1
x
x
x v v
x
Differentiate x withrespect to x
sol Let y x
and z x
Differentiating and withrespect to x we get
dy d d d
x x x u u v u
dx dx dx dx
x x
x
x x










 
 
 
 
 
 
 

 
 
1
1
2
1
sin
2
2
2 1
sin
1
1
,
sin ln
1
1
1
1 .sin
ln .
x
x
dz
and
dx x
dy
dy dx
Now
dz
dz
dx
x x
x
x x
x
x x
x x Ans
x







 

 

 


 

 
 
 
 
2
1 1
2
1
2
1
1
2
1
1
1
1 1
7. tan tan .
1 1
: , tan
sin
1 sin 1
tan ;
sin sin
cos 1
tan
sin
cos 1
tan
sin
1 cos
tan
x
Differentiate withrespect to x
x
x
sol Let y
x
putting x
x


 




 






 
 
 
 
 
 
 
  
 
 
  
 
 
    
   
 
 
 

  
 
 
 

  
 

 
 
2
1
1
1
1
1
1
sin
2sin
2
tan
2sin cos
2 2
sin
2
tan
cos
2
tan tan
2
tan tan
2
2
1
sin (1)
2
tan (2)
(1) (2) ,
x
and z x
Differentiating and withrespect to x we get
dy
d



 














 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

 
2
2
2
2
2
2
1 1
1
2 1
,
1
2 1
1
1
1
.
2 1
dz
and
x dx x
x
dy
dy dx
Now
dz
dz
dx
x
x
x
Ans
x
  








 


13. 13
 
1 1
2 2
1
2
1
2 1
1
1
1
1
2
1
8. sec tan .
2 1 1
1
: , sec
2 1
cos
1
sec ;
2cos 1 cos
1
sec
cos2
sec sec2
2
2cos (1)
tan
1
x
Differentiate withrespectto
x x
sol Let y
x
putting x
x
x
x
and z
x

 



 







 
 
 
 

  
 
 
  

 

 
 
  
 
  
   
 
  
 






1
1
2
1
2
1
1
1
2 2
sin
sin
tan ;
sin
1 sin
sin
tan
cos
sin
tan
cos
tan .tan
sin (2)
(1) (2) ,
2 1
1 1
,
putting x
x
x
Differentiating and withrespectto x we get
dy dz
and
dx dx
x x
dy
Now
dz

















 
 

   
    
 
  
 
 
  
 
 
  
 



  
 

 
2
2
2
1
1
1
2 .
dy
dx
dz
dx
x
x
Ans




 

14. 14
Homework:-
1.
2
1 1
1 1
tan tan .
x
Differentiate withrespect to x
x
 
 
 
 
 
 
Ans:
1
2
2.
1
sin
cos3 .
x
Differentiate e withrespectto x

Ans:
1
sin
2
3 1 .sin3
x
e
x x



3.
2
1 1
2 2
1 2
cos tan .
1 1
x x
Differentiate withrespectto
x x
 
 
  
   
 
 
 
Ans: 1
Successive derivative
Successive derivative: If  
y f x
 be a function of x then the first order derivative of y with
respect to x is denoted by      
   
1 1
' '
1
, , , , , .
x
dy
f x y y f x f x etc
dx
Again the derivative of first ordered derivative of y with respect to x is called second order derivative
and is denoted by      
   
2
2 2
'' ''
2
2
, , , , , .
x
d y
f x y y f x f x etc
dx
Similarly, the nth derivative of y with respect to x is denoted by
     
   
, , , , , .
n
n n
n n
n x
n
d y
f x y y f x f x etc
dx
 Find the nth derivative of the following functions:
 
  
    
 
    
 
  
1
1
2
2
3
3
1.
: ,
,
1
1 2
,
1 2 1 ; ,
1 2 1
1 2 3.2.1
! .
n
n
n
n
n
n r
r
n n
n
y x
sol Giventhat y x
Differentiating withrespectto xwe get
y nx
y n n x
y n n n x
Similarly
y n n n n r x where r n
y n n n n n x
n n n
n Ans








  
   
     
     
  

1
2
2
3
3
2.
: ,
,
,
; ,
.
ax
ax
ax
ax
ax
r ax
r
n ax
n
y e
sol Giventhat y e
Differentiating withrespectto x we get
y ae
y a e
y a e
Similarly
y a e where r n
y a e Ans



 
 
 
 

15. 15
 
 
 
  
   
    
  
    
  
 
 
1
1
2
2
2
3
3
3
3.
: ,
,
1
1 2
,
1 2 1 ; ,
1 2 1
!
!
m
m
m
m
m
m r
r
r
m n
n
n
n
y ax b
sol Giventhat y ax b
Differentiating withrespect to x we get
y am ax b
y a m m ax b
y a m m m ax b
Similarly
y a m m m m r ax b where r n
y a m m m m n ax b
m
a ax b
m n





 
 
 
   
    
      
      
 

.
m n
Ans

 
 
 
 
 
 
 
 
1
2
2
2
2
3
3
3
4. sin
: , sin
,
cos
sin
2
cos
2
sin
2 2
2
sin
2
2
cos
2
2
sin
2 2
y ax b
sol Giventhat y ax b
Differentiating withrespect to x we get
y a ax b
a ax b
y a ax b
a ax b
a ax b
y a ax b
a ax


 


 
 
 
 
 
  
 
 
 
   
 
 
 
   
 
 
 
  
 
 
 
   
 
 
    
 
 
 
3 3
sin
2
,
sin ; ,
2
sin .
2
r
r
n
n
b
a ax b
Similarly
r
y a ax b where r n
n
y a ax b Ans



 

 
 
 
  
 
 
 
   
 
 
 
   
 
 
 
 
 
 
 
 
 
 
1
2
2
2
2
3
3
3
5. cos
: , cos
,
sin
cos
2
sin
2
cos
2 2
2
cos
2
2
sin
2
2
cos
2 2
y ax b
sol Giventhat y ax b
Differentiating withrespect to x we get
y a ax b
a ax b
y a ax b
a ax b
a ax b
y a ax b
a


 


 
 
 
  
 
  
 
 
 
    
 
 
 
   
 
 
 
  
 
 
 
    
 
 
   
 
 
 
3 3
cos
2
,
cos ; ,
2
cos .
2
r
r
n
n
ax b
a ax b
Similarly
r
y a ax b where r n
n
y a ax b Ans



 
 
 
 
 
  
 
 
 
   
 
 
 
   
 
 

16. 16
 
 
   
   
 
 
1
2 2 1
1
6. sin
: , sin
,
sin cos
sin cos
cos sin
tan
, cos sin sin cos
ax
ax
ax ax
ax
ax
y e bx c
sol Giventhat y e bx c
Differentiating withrespect to x we get
y ae bx c be bx c
e a bx c b bx c
put a r and b r
b
r a b and
a
Now y e r bx c r b
 

 

 
 
   
   
 
 
     
 
    
 
 
   
 
   
 
 
 
 
 
2
2
3
3
2 2 1
sin
sin cos
cos sin sin cos
sin 2
sin 3
,
sin
sin tan .
ax
ax
ax
ax
ax
n ax
n
n
ax
x c
re bx c
y re a bx c b bx c
re r bx c r bx c
r e bx c
y r e bx c
Similarly
y r e bx c n
b
a b e bx c n Ans
a

 
   





  
      
     
  
   
  
 
 
     
 
 
 
 
 
 
 
 
 
   
 
1
2
2 2
3
3 3
4
4 4
1
7. ln
: , ln
,
1.
1.2
1.2.3
,
1 1 !
.
n n
n n
y ax b
sol Giventhat y ax b
Differentiating withrespect to x we get
a
y
ax b
a
y
ax b
a
y
ax b
a
y
ax b
Similarly
n a
y Ans
ax b

 
 


  

 

  

 
 

 
1
2
1
2 2
2
8. sin cos 1 1 sin 2 .
: , sin cos
,
cos sin
sin cos
2 2
cos sin
2 2
r
r
r
If y nx nx then showthat y n nx
sol Giventhat y nx nx
Differentiating with respect to x we get
y n nx n nx
n nx n nx
y n nx n nx
 
 
 
    
 
 
 
   
   
   
   
   
    
   
   
 2 2
3 3
3
3 3
3
2
2 2
sin cos
2 2
2 2
cos sin
2 2
3 3
sin cos
2 2
,
sin cos
2 2
sin cos
2 2
r
r
r
n nx n nx
y n nx n nx
n nx n nx
Similarly
r r
y n nx n nx
r r
n nx nx
 
 
 
 
 
   
  
   
   
   
    
   
   
   
   
   
   
   
   
   
   
 
   
   
 
   
   
 
 
 
1
2
1
2
2 2
1
2
1
2
1
2
sin cos 2sin cos
2 2 2 2
1 sin 2
2
1 sin 2
1 1 sin 2 .
r
r
r
r
r
r r r r
n nx nx nx nx
r
n nx
n r nx
n nx showed
   



 
 
 
 
       
      
       
 
       
 
 
 
  
 
 
 
 
  
 
 
 
  
 

17. 17
Homework:-
1. Find the nth derivative of the following functions:
a. 2
1
5 6
y
x x

 
Ans:  
   
1 1
1 1
1 !
2 3
n
n n n
y n
x x
 
 
  
 
 
 
 
b. 2
2 3
.
3 2
x
y
x x


 
Ans:  
   
1 1
1 1
1 !
1 2
n
n n n
y n
x x
 
 
  
 
 
 
 
   
2
2
1 1
2 2
1
3 3
2
2 3 2 3
1
9.
3 2
1
: ,
3 2
1 1
2 1
( 2) ( 1)
,
( 1)( 2) ( 1)( 1)
( 1)( 2)( 2) ( 1)( 2)( 1)
( 1) 2!( 2) ( 1) 2!( 1)
,
y
x x
sol Giventhat y
x x
x x
x x
Differentiating withrespect to x we get
y x x
y x x
x x
Similarly
y
 
 
 
 

 

 
 
 
   
     
        
     
    
 
   
( 1) ( 1)
1 1
1 !( 2) 1 !( 1)
1 1
1 !
2 1
n n
n n
n
n
n n
n x n x
n Ans
x x
   
 
     
 
 
  
 
 
 
 