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Short notes on Differentiation | Engineering Mathematics - Civil Engineering (CE) PDF Download

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 Page 1


    
 
Differentiability 
Let a real valued function f(x) be defined on an I and let x
0
 be a point in I. Then, if 
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
 
exists and is equal to ?, then f(x) is said to be differentiable at x
0
 and ? is called the 
derivative of f(x) at x = x
0
. 
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be 
differentiable in (a, b). 
 
Properties of differentiation 
? Let the functions f and g be differentiable at a point x
0
. Then 
 i) (cf ?) x
0
 = cf ?(x
0
), c any constant. 
 ii) (f ? g) ?x
0
 = f ?(x
0
) ? g ?(x
0
) 
 iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
) 
 iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
 
? If f is differentiable at x
0
 and g is differentiable at f(x
0
) then the composite function 
 h = g(f(x)) is differentiable at x
0
 and 
 h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
) 
? If the function y = f(x) is represented in the parametric form as 
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then    
 f ?(x) = 
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
 
 
 
Note :  If a function is differentiable at x = x
0
, then it is continuous at x = x
0
. 
 However the converse need not be true. 
 
 
 
 
 
Page 2


    
 
Differentiability 
Let a real valued function f(x) be defined on an I and let x
0
 be a point in I. Then, if 
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
 
exists and is equal to ?, then f(x) is said to be differentiable at x
0
 and ? is called the 
derivative of f(x) at x = x
0
. 
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be 
differentiable in (a, b). 
 
Properties of differentiation 
? Let the functions f and g be differentiable at a point x
0
. Then 
 i) (cf ?) x
0
 = cf ?(x
0
), c any constant. 
 ii) (f ? g) ?x
0
 = f ?(x
0
) ? g ?(x
0
) 
 iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
) 
 iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
 
? If f is differentiable at x
0
 and g is differentiable at f(x
0
) then the composite function 
 h = g(f(x)) is differentiable at x
0
 and 
 h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
) 
? If the function y = f(x) is represented in the parametric form as 
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then    
 f ?(x) = 
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
 
 
 
Note :  If a function is differentiable at x = x
0
, then it is continuous at x = x
0
. 
 However the converse need not be true. 
 
 
 
 
 
   
Standard formulae 
 
Function Derivative Function Derivative 
k(constant) 0 
1
x
sin
a
?
 
22
1
ax ?
 
x
n
 nx
n ?1 1
x
cos
a
?
 
22
1
ax
?
?
 
logx 1/x 
1
x
tan
a
?
 
22
a
ax ?
 
e
x 
e
x 1
x
cot
a
?
 
22
a
ax
?
?
 
a
x 
a
x
loga 
1
x
sec
a
?
 
22
a
xx a ?
 
sin x cos x 
1
x
cosec
a
?
 
22
a
xx a
?
?
 
cos x ? sin x sin h x cos h x 
tan x sec
2
x cos h x sin h x 
cot x ? cosec
2
 x sin h
?1
 x 
2
1
x1 ?
 
sec x sec x tan x cos h
?1
 x 
2
1
x1 ?
 
cosec x ? cosec x cot x tan h
?1
 x 
2
1
x1 ?
 
 
Solved Example 8 : 
Show that the function 
f(x) = 
2
xcos(1/x), x 0
0x0
? ?
?
?
?
 
is differentiable at x = 0 but f ?(x) is not 
continuous at x = 0. 
Solution : 
We have 
x0
lim f(x) 0 f(0)
?
? ? 
? f(x) is continuous at x = 0  
Now f ?(0) = 
?
?
x0
f(x) f(0)
lim
x
 
  
?
?? ??
? ?
?? ??
?? ??
x0
1
lim xcos 0
x
 
Hence f(x) is differentiable at  
x = 0  and f ?(0) = 0  
for x ? 0  we have 
Page 3


    
 
Differentiability 
Let a real valued function f(x) be defined on an I and let x
0
 be a point in I. Then, if 
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
 
exists and is equal to ?, then f(x) is said to be differentiable at x
0
 and ? is called the 
derivative of f(x) at x = x
0
. 
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be 
differentiable in (a, b). 
 
Properties of differentiation 
? Let the functions f and g be differentiable at a point x
0
. Then 
 i) (cf ?) x
0
 = cf ?(x
0
), c any constant. 
 ii) (f ? g) ?x
0
 = f ?(x
0
) ? g ?(x
0
) 
 iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
) 
 iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
 
? If f is differentiable at x
0
 and g is differentiable at f(x
0
) then the composite function 
 h = g(f(x)) is differentiable at x
0
 and 
 h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
) 
? If the function y = f(x) is represented in the parametric form as 
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then    
 f ?(x) = 
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
 
 
 
Note :  If a function is differentiable at x = x
0
, then it is continuous at x = x
0
. 
 However the converse need not be true. 
 
 
 
 
 
   
Standard formulae 
 
Function Derivative Function Derivative 
k(constant) 0 
1
x
sin
a
?
 
22
1
ax ?
 
x
n
 nx
n ?1 1
x
cos
a
?
 
22
1
ax
?
?
 
logx 1/x 
1
x
tan
a
?
 
22
a
ax ?
 
e
x 
e
x 1
x
cot
a
?
 
22
a
ax
?
?
 
a
x 
a
x
loga 
1
x
sec
a
?
 
22
a
xx a ?
 
sin x cos x 
1
x
cosec
a
?
 
22
a
xx a
?
?
 
cos x ? sin x sin h x cos h x 
tan x sec
2
x cos h x sin h x 
cot x ? cosec
2
 x sin h
?1
 x 
2
1
x1 ?
 
sec x sec x tan x cos h
?1
 x 
2
1
x1 ?
 
cosec x ? cosec x cot x tan h
?1
 x 
2
1
x1 ?
 
 
Solved Example 8 : 
Show that the function 
f(x) = 
2
xcos(1/x), x 0
0x0
? ?
?
?
?
 
is differentiable at x = 0 but f ?(x) is not 
continuous at x = 0. 
Solution : 
We have 
x0
lim f(x) 0 f(0)
?
? ? 
? f(x) is continuous at x = 0  
Now f ?(0) = 
?
?
x0
f(x) f(0)
lim
x
 
  
?
?? ??
? ?
?? ??
?? ??
x0
1
lim xcos 0
x
 
Hence f(x) is differentiable at  
x = 0  and f ?(0) = 0  
for x ? 0  we have 
    
f ?(x) = 
?? ?? ?? ? ?
?? ?
?? ?? ?? ? ?
?? ?? ? ? ??
2
2
111
2xcos x sin
xx x
 
?? ? ?
??
?? ? ?
?? ? ?
11
2xcos sin
xx
 
Now ??
x0
lim f x
?
? does not exist as 
x0
1
lim sin
x
?
??
??
??
 
does not exist. Therefore, f ?(x) is not 
continuous at x = 0. 
 
Solved Example 9 : 
3
yx ? is defined and continuous for all x 
investigate whether this function has a 
derivative at x = 0. 
Solution : 
?y = 
33
xx x ?? ? at x = 0 
 
?y = 
3
x ? 
 
? 
??
3
2
3
yx 1
xx
x
??
??
??
?
 
? 
??
2 x0 x0
3
y1
lim lim
x
x
?? ??
?
????
?
?
 
? There is no finite derivative. 
 
Solved Example 10 : 
Investigate the function f(x) = | x | for 
differentiability at the point x = 0  
 
Solution : 
 
y = f(x) = | x | 
?y = | x + ?x | ? | ?x | 
 
At x = 0, we have ?y = | ?x | 
? 
x
y
xx
?
?
?
? ?
 
x0
y
lim 1
x
?
??
?
? ?
?
  
x0
y
lim 1
x
?
??
?
?
?
 
Since the right and the left side derivative 
are not equal, the function f(x) = |x| is not 
differentiable at the point zero. 
 
Solved Example 11 : 
Show that 
??
nn1
d
xnx
dx
?
? 
Solution : 
Let y = x
n
 
Let x receive a small increment ?x and let 
the corresponding increment in y be ?y 
then  
 y + ?y = (x + ?x)
n
 
By subtraction, ?y = (x + ?x)
n
 ? x
n
 
? 
?? ??
nn
nn
xx x xx x
y
xx xxx
?? ? ?? ?
?
??
? ????
 
? 
??
n
n
n1
x0 x x x
xx x
y
lim lim nx
xxxx
?
?? ???
?? ?
?
??
? ?? ?
 
? 
??
n1 n n1
dy d
nx or x nx
dx dx
? ?
??
 
 
Page 4


    
 
Differentiability 
Let a real valued function f(x) be defined on an I and let x
0
 be a point in I. Then, if 
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
 
exists and is equal to ?, then f(x) is said to be differentiable at x
0
 and ? is called the 
derivative of f(x) at x = x
0
. 
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be 
differentiable in (a, b). 
 
Properties of differentiation 
? Let the functions f and g be differentiable at a point x
0
. Then 
 i) (cf ?) x
0
 = cf ?(x
0
), c any constant. 
 ii) (f ? g) ?x
0
 = f ?(x
0
) ? g ?(x
0
) 
 iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
) 
 iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
 
? If f is differentiable at x
0
 and g is differentiable at f(x
0
) then the composite function 
 h = g(f(x)) is differentiable at x
0
 and 
 h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
) 
? If the function y = f(x) is represented in the parametric form as 
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then    
 f ?(x) = 
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
 
 
 
Note :  If a function is differentiable at x = x
0
, then it is continuous at x = x
0
. 
 However the converse need not be true. 
 
 
 
 
 
   
Standard formulae 
 
Function Derivative Function Derivative 
k(constant) 0 
1
x
sin
a
?
 
22
1
ax ?
 
x
n
 nx
n ?1 1
x
cos
a
?
 
22
1
ax
?
?
 
logx 1/x 
1
x
tan
a
?
 
22
a
ax ?
 
e
x 
e
x 1
x
cot
a
?
 
22
a
ax
?
?
 
a
x 
a
x
loga 
1
x
sec
a
?
 
22
a
xx a ?
 
sin x cos x 
1
x
cosec
a
?
 
22
a
xx a
?
?
 
cos x ? sin x sin h x cos h x 
tan x sec
2
x cos h x sin h x 
cot x ? cosec
2
 x sin h
?1
 x 
2
1
x1 ?
 
sec x sec x tan x cos h
?1
 x 
2
1
x1 ?
 
cosec x ? cosec x cot x tan h
?1
 x 
2
1
x1 ?
 
 
Solved Example 8 : 
Show that the function 
f(x) = 
2
xcos(1/x), x 0
0x0
? ?
?
?
?
 
is differentiable at x = 0 but f ?(x) is not 
continuous at x = 0. 
Solution : 
We have 
x0
lim f(x) 0 f(0)
?
? ? 
? f(x) is continuous at x = 0  
Now f ?(0) = 
?
?
x0
f(x) f(0)
lim
x
 
  
?
?? ??
? ?
?? ??
?? ??
x0
1
lim xcos 0
x
 
Hence f(x) is differentiable at  
x = 0  and f ?(0) = 0  
for x ? 0  we have 
    
f ?(x) = 
?? ?? ?? ? ?
?? ?
?? ?? ?? ? ?
?? ?? ? ? ??
2
2
111
2xcos x sin
xx x
 
?? ? ?
??
?? ? ?
?? ? ?
11
2xcos sin
xx
 
Now ??
x0
lim f x
?
? does not exist as 
x0
1
lim sin
x
?
??
??
??
 
does not exist. Therefore, f ?(x) is not 
continuous at x = 0. 
 
Solved Example 9 : 
3
yx ? is defined and continuous for all x 
investigate whether this function has a 
derivative at x = 0. 
Solution : 
?y = 
33
xx x ?? ? at x = 0 
 
?y = 
3
x ? 
 
? 
??
3
2
3
yx 1
xx
x
??
??
??
?
 
? 
??
2 x0 x0
3
y1
lim lim
x
x
?? ??
?
????
?
?
 
? There is no finite derivative. 
 
Solved Example 10 : 
Investigate the function f(x) = | x | for 
differentiability at the point x = 0  
 
Solution : 
 
y = f(x) = | x | 
?y = | x + ?x | ? | ?x | 
 
At x = 0, we have ?y = | ?x | 
? 
x
y
xx
?
?
?
? ?
 
x0
y
lim 1
x
?
??
?
? ?
?
  
x0
y
lim 1
x
?
??
?
?
?
 
Since the right and the left side derivative 
are not equal, the function f(x) = |x| is not 
differentiable at the point zero. 
 
Solved Example 11 : 
Show that 
??
nn1
d
xnx
dx
?
? 
Solution : 
Let y = x
n
 
Let x receive a small increment ?x and let 
the corresponding increment in y be ?y 
then  
 y + ?y = (x + ?x)
n
 
By subtraction, ?y = (x + ?x)
n
 ? x
n
 
? 
?? ??
nn
nn
xx x xx x
y
xx xxx
?? ? ?? ?
?
??
? ????
 
? 
??
n
n
n1
x0 x x x
xx x
y
lim lim nx
xxxx
?
?? ???
?? ?
?
??
? ?? ?
 
? 
??
n1 n n1
dy d
nx or x nx
dx dx
? ?
??
 
 
   
 Mean Value Theorems 
Rolle's Theorem 
The theorem states that 
? if f(x) is continuous in the closed interval [a, b] and 
? if f ?(x) exists in open interval (a, b) and 
? if f(a) = f(b) 
Then there exists at least one value c in (a, b) such that f ?(c) = 0 
 
Geometric Interpretation 
 
 
 
 
 
 
 
 
 
There exists at last one point at which slope of the tangent is 0 or the tangent is parallel 
to x-axis 
 
Lagrange's Mean Value Theorem 
The theorem states that 
? if f(x) is continuous in the closed interval [a, b] and 
? if f ?(x) exists in open interval (a, b) 
Then there exists one value c such that  
 
f ?(c) = 
??
??
f(b) f(a)
ba
?
?
 
 
 
 
 
 
 
 
 
 
a c b x 
y 
f(a) = f(b) 
y = f(x) 
f ?(c) = 0 
Page 5


    
 
Differentiability 
Let a real valued function f(x) be defined on an I and let x
0
 be a point in I. Then, if 
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
 
exists and is equal to ?, then f(x) is said to be differentiable at x
0
 and ? is called the 
derivative of f(x) at x = x
0
. 
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be 
differentiable in (a, b). 
 
Properties of differentiation 
? Let the functions f and g be differentiable at a point x
0
. Then 
 i) (cf ?) x
0
 = cf ?(x
0
), c any constant. 
 ii) (f ? g) ?x
0
 = f ?(x
0
) ? g ?(x
0
) 
 iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
) 
 iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
 
? If f is differentiable at x
0
 and g is differentiable at f(x
0
) then the composite function 
 h = g(f(x)) is differentiable at x
0
 and 
 h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
) 
? If the function y = f(x) is represented in the parametric form as 
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then    
 f ?(x) = 
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
 
 
 
Note :  If a function is differentiable at x = x
0
, then it is continuous at x = x
0
. 
 However the converse need not be true. 
 
 
 
 
 
   
Standard formulae 
 
Function Derivative Function Derivative 
k(constant) 0 
1
x
sin
a
?
 
22
1
ax ?
 
x
n
 nx
n ?1 1
x
cos
a
?
 
22
1
ax
?
?
 
logx 1/x 
1
x
tan
a
?
 
22
a
ax ?
 
e
x 
e
x 1
x
cot
a
?
 
22
a
ax
?
?
 
a
x 
a
x
loga 
1
x
sec
a
?
 
22
a
xx a ?
 
sin x cos x 
1
x
cosec
a
?
 
22
a
xx a
?
?
 
cos x ? sin x sin h x cos h x 
tan x sec
2
x cos h x sin h x 
cot x ? cosec
2
 x sin h
?1
 x 
2
1
x1 ?
 
sec x sec x tan x cos h
?1
 x 
2
1
x1 ?
 
cosec x ? cosec x cot x tan h
?1
 x 
2
1
x1 ?
 
 
Solved Example 8 : 
Show that the function 
f(x) = 
2
xcos(1/x), x 0
0x0
? ?
?
?
?
 
is differentiable at x = 0 but f ?(x) is not 
continuous at x = 0. 
Solution : 
We have 
x0
lim f(x) 0 f(0)
?
? ? 
? f(x) is continuous at x = 0  
Now f ?(0) = 
?
?
x0
f(x) f(0)
lim
x
 
  
?
?? ??
? ?
?? ??
?? ??
x0
1
lim xcos 0
x
 
Hence f(x) is differentiable at  
x = 0  and f ?(0) = 0  
for x ? 0  we have 
    
f ?(x) = 
?? ?? ?? ? ?
?? ?
?? ?? ?? ? ?
?? ?? ? ? ??
2
2
111
2xcos x sin
xx x
 
?? ? ?
??
?? ? ?
?? ? ?
11
2xcos sin
xx
 
Now ??
x0
lim f x
?
? does not exist as 
x0
1
lim sin
x
?
??
??
??
 
does not exist. Therefore, f ?(x) is not 
continuous at x = 0. 
 
Solved Example 9 : 
3
yx ? is defined and continuous for all x 
investigate whether this function has a 
derivative at x = 0. 
Solution : 
?y = 
33
xx x ?? ? at x = 0 
 
?y = 
3
x ? 
 
? 
??
3
2
3
yx 1
xx
x
??
??
??
?
 
? 
??
2 x0 x0
3
y1
lim lim
x
x
?? ??
?
????
?
?
 
? There is no finite derivative. 
 
Solved Example 10 : 
Investigate the function f(x) = | x | for 
differentiability at the point x = 0  
 
Solution : 
 
y = f(x) = | x | 
?y = | x + ?x | ? | ?x | 
 
At x = 0, we have ?y = | ?x | 
? 
x
y
xx
?
?
?
? ?
 
x0
y
lim 1
x
?
??
?
? ?
?
  
x0
y
lim 1
x
?
??
?
?
?
 
Since the right and the left side derivative 
are not equal, the function f(x) = |x| is not 
differentiable at the point zero. 
 
Solved Example 11 : 
Show that 
??
nn1
d
xnx
dx
?
? 
Solution : 
Let y = x
n
 
Let x receive a small increment ?x and let 
the corresponding increment in y be ?y 
then  
 y + ?y = (x + ?x)
n
 
By subtraction, ?y = (x + ?x)
n
 ? x
n
 
? 
?? ??
nn
nn
xx x xx x
y
xx xxx
?? ? ?? ?
?
??
? ????
 
? 
??
n
n
n1
x0 x x x
xx x
y
lim lim nx
xxxx
?
?? ???
?? ?
?
??
? ?? ?
 
? 
??
n1 n n1
dy d
nx or x nx
dx dx
? ?
??
 
 
   
 Mean Value Theorems 
Rolle's Theorem 
The theorem states that 
? if f(x) is continuous in the closed interval [a, b] and 
? if f ?(x) exists in open interval (a, b) and 
? if f(a) = f(b) 
Then there exists at least one value c in (a, b) such that f ?(c) = 0 
 
Geometric Interpretation 
 
 
 
 
 
 
 
 
 
There exists at last one point at which slope of the tangent is 0 or the tangent is parallel 
to x-axis 
 
Lagrange's Mean Value Theorem 
The theorem states that 
? if f(x) is continuous in the closed interval [a, b] and 
? if f ?(x) exists in open interval (a, b) 
Then there exists one value c such that  
 
f ?(c) = 
??
??
f(b) f(a)
ba
?
?
 
 
 
 
 
 
 
 
 
 
a c b x 
y 
f(a) = f(b) 
y = f(x) 
f ?(c) = 0 
    
Geometric Interpretation 
 
 
 
 
 
 
 
 
 
 
 
 
 
There exists at last one point at which the tangent is parallel to the secant through the 
end points. 
 
Cauchy's Mean Value Theorem 
The theorem states that 
? if f(x) and g(x) are both continuous in closed interval [a, b] and 
? if f ?(x) and g(x) both exist in open interval (a, b) 
Then there exists at least one value c such that 
 
??
?? ??
??
f(c) f(b) f(a)
g(c) g(b) g(a)
? ?
?
? ?
 
 
 
 
Maxima and Minima 
A function f(x) is said to have a maximum value at x = a if f(a) is larger than any other 
values of f(x) in the immediate neighbourhood of ‘a’.  It has a minimum value if f(a) is less 
than any other value of f(x) sufficiently near ‘a’. 
 
Finding Maximum and Minimum values of y = f(x) 
? Get 
dy
dx
 and
2
2
dy
dx
. Solve 
dy
dx
 = 0 and consider its roots. These are the values of x 
which make 
dy
dx
= 0. 
? For each of these values of x, calculate the corresponding value of y and examine 
the sign of
2
2
dy
dx
. 
? If the sign is ?ve the corresponding value of y is a maximum. 
 If the sign is +ve, the corresponding value of y is minimum. 
a c 
b 
f(a) 
f(b) 
f ?(c) = 
f(b) f(a)
ba
?? ?
?
 
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FAQs on Short notes on Differentiation - Engineering Mathematics - Civil Engineering (CE)

1. What is the main purpose of differentiation in mathematics?
Ans. The main purpose of differentiation in mathematics is to find the rate at which a function is changing at a particular point.
2. How is differentiation used in real-life applications?
Ans. Differentiation is used in various real-life applications such as calculating the speed of a moving object, determining the slope of a curve, and optimizing functions in economics and engineering.
3. What are the basic rules of differentiation that one should know?
Ans. Some basic rules of differentiation include the power rule, product rule, quotient rule, and chain rule, which are essential for finding the derivative of functions.
4. How does differentiation relate to integration in calculus?
Ans. Differentiation and integration are inverse processes in calculus. Differentiation finds the rate of change, while integration finds the accumulation of change. The Fundamental Theorem of Calculus establishes a connection between the two.
5. Can differentiation be used to find maximum and minimum points of a function?
Ans. Yes, differentiation can be used to find maximum and minimum points of a function by setting the derivative equal to zero and solving for critical points. The first derivative test or second derivative test can then be applied to determine the nature of these points.
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