UNIT 1
Differential Calculus
Continuous Function
A function f(x) is said to be continuous at x = a if
f(a) exist is function must exist at x = a
Differentiation
Let y = f(x) be any function A function f(x) is said to be differentiable at x = a if
is exist and it is denoted by
Note:
A differentiable function is always continuous but converse need not be true.
If
f(x) is continuous in the closed [a, b]
f(x) is differentiable in (a, b) &
f(a) = f(b)
Then there exist at least one value ‘c’ in (a, b) such that f’(c) = 0.
Q1)Verify Rolle’s theorem for the function f(x) = x2 for
S1)
Here f(x) = x2; 
Since f(x) is algebraic polynomial which is continuous in [1, 1] 
Consider f(x) = x2 
a) Diff. w.r.t. x we get 
b) f'(x) = 2x 
Clearly f’(x) exist in (1, 1) and does not becomes infinite. 
Clearly 
f(1) = (1)2 = 1 
f(1) = (1)2 = 1 
f(1) = f(1). 
Hence by Rolle’s theorem, there exist such that 
f’(c) = 0 
i.e. 2c = 0 
c = 0 
Thus such that 
f'(c) = 0 
Hence Rolle’s Theorem is verified. 

Q2) Verify Rolle’s Theorem for the function f(x) = ex(sin x – cos x) in
S2)
Here f(x) = ex(sin x – cos x); 
Clearly ex is an exponential function continuous for every also sin x and cos x are Trigonometric functions Hence (sin x – cos x) is continuous in and Hence ex(sin x – cos x) is continuous in . 
Consider 
f(x) = ex(sin x – cos x) 
diff. w.r.t. x we get 
f’(x) = ex(cos x + sin x) + ex(sin x + cos x) 
= ex[2sin x] 
Clearly f’(x) is exist for each & f’(x) is not infinite. 
Hence f(x) is differentiable in . 
Consider 
Also, 
Thus 
Hence all the conditions of Rolle’s theorem are satisfied, so there exist such, that 
i.e. 
i.e. sin c = 0 
But 
Hence Rolle’s theorem is verified. 
Q3) Verify whether Rolle’s theorem is applicable or not for
S3)
Here f(x) = x2; 
Clearly x2 is an algebraic polynomial hence it is continuous in [2, 3] 
Consider 
Clearly f’(x) is exist for each 
Consider 
Thus . 
Thus all conditions of Rolle’s theorem are not satisfied Hence Rolle’s theorem is not applicable for f(x) = x2 in [2, 3] 
Q4)
a) Show that between any two real root of equation , is at least one real root of .
b) Discuss the applicability of Rolle’s theorem for the function
Lagrange’s Mean value Theorem:
Statement: If
f(x) is continuous in [a, b]
f(x) is differentiable in (a, b) then there exist at least one value such that
Q5) Verify the Lagrange’s mean value theorem for
S5)
Here 
Clearly f(x) = log x is logarithmic function. Hence it is continuous in [1, e] 
Consider f(x) = log x. 
Diff. w.r.t. x we get, 
Clearly f’(x) is exist for each value of & is finite. 
Hence all conditions of LMVT are satisfied Hence at least 
Such that 

i.e. 
i.e. 
i.e. 
i.e. 
since e = 2.7183 
Clearly c = 1.7183 
Hence LMVT is verified. 


Q6) Verify mean value theorem for f(x) = tan1x in [0, 1]
S6)
Here ; 
Clearly is an inverse trigonometric function and hence it is continuous in [0, 1] 
Consider 
diff. w.r.t. x we get, 
Clearly f’(x) is continuous and differentiable in (0, 1) & is finite 
Hence all conditions of LMVT are satisfied, Thus there exist 
Such that 
i.e. 
i.e. 
i.e. 
i.e. 
Clearly 
Hence LMVT is verified. 

Meaning of sign of Derivative: 
Let f(x) satisfied LMVT in [a, b] 
Let x1 and x2 be any two points laying (a, b) such that x1 < x2 
Hence by LMVT, such that 
i.e. … (1) 
Cast I: 
If then 
i.e. 
is constant function 
Case II: 
If then from equation (1) 
i.e. 
means x2  x1 > 0 and 
Thus for x2 > x1 
Thus f(x) is increasing function is (a, b) 
Case III: 
If 
Then from equation (1) 
i.e. 
since and then hence f(x) is strictly decreasing function. 

Q7) Prove that
And hence show that
S7)
Let ; 
Clearly is an logarithmic function and hence it is continuous also 
Consider 
diff. w.r.t. x we get, 
Clearly f’(x) exist and finite in (a, b) Hence f(x) is continuous and differentiable in (a, b). Hence by LMVT 
Such that 
i.e. 
i.e. 
since 
a < c < b 
i.e. 
i.e. 
i.e. 
i.e. 
Hence the result 
Now put a = 5, b = 6 we get 
Hence the result 
Q8) Prove that , use mean value theorem to prove that,
Hence show that
S8)
Let f(x) = sin1x; 
Clearly f(x) is inverse trigonometric function and hence it is continuous in [a, b] 
Consider f(x) = sin1x 
diff. w.r.t. x we get, 
Clearly f’(x) is finite and exist for . Hence by LMVT, such that 
i.e. 
since a < c < b 
i.e. 
i.e. 
i.e. 
i.e. 
Hence the result 
Put we get 
i.e. 
i.e. 
i.e. 
i.e. 
Hence the result 
Cauchy’s Mean Value Theorem:
Statement:
If f(x) and g(x) are any two functions such that
f(x) and g(x) are continuous in (a, b)
both f(x) and g(x) are derivable in (a, b)
Then for any value of , at least such that
Q9)
Verify Cauchy mean value theorems for & in
S9)
Let & ; 
Clearly f(x) and g(x) both are trigonometric functions. Hence continuous in 
Since & 
diff. w.r.t. x we get, 
& 
Clearly both f’(x) and g’(x) exist & finite in . Hence f(x) and g(x) is derivable in and 

Hence by Cauchy mean value theorem, there exist at least such that 
i.e. 
i.e. 1 = cot c 
i.e. 
clearly 
Hence Cauchy mean value theorem is verified. 

Q10) Considering the functions ex and ex, show that c is arithmetic mean of a & b.
Solution:
Clearly f(x) and g(x) are exponential functions Hence they are continuous in [a, b]. 
Consider & 
diff. w.r.t. x we get 
and 
Clearly f(x) and g(x) are derivable in (a, b) 
By Cauchy’s mean value theorem such that 
i.e. 
i.e. 
i.e. 
i.e. 
i.e. 
i.e. 
Thus 
i.e. c is arithmetic mean of a & b. 
Hence the result 

Q11) Show that
Prove that if
and Hence show that
Verify Cauchy’s mean value theorem for the function x2 and x4 in [a, b] where a, b > 0
If for then prove that,
[Hint:, ]
Expansions of functions
In this topic we learn two important series expansions namely
a) Maclaurin’s series
b) Taylors Series
Maclaurin’s Series Expansions
Statement:
Maclaurin’s series of f(x) at x = 0 is given by,
Expansion of some standard functions
1) f(x) = ex then
Proof:
Here 




By Maclaurin’s series we get, 
i.e. 
Note that 
1) Replace x by –x we get 
2) f(x) = sin x then
Proof: 
Let (x) = sin x 
Then by Maclaurin’s series, 
… (1) 
Since 





By equation (i) we get, 
3)
Then
Proof: 
Let f(x) = cos x 
Then by Maclaurin’s series, 
… (1) 
Since 






From Equation (1) 
4) then
Proof: 
Here f(x) = tan x 
By Maclaurin’s expansion, 
… (1) 
Since 



….. 
By equation (1) 
5) Then
Proof: 
Here f(x) = sin hx. 
By Maclaurin’s expansion, 
(1) 




By equation (1) we get, 
6) . Then
Proof: 
Here f(x) = cos hx 
By Maclaurin’s expansion 
(1) 




By equation (1) 
7) f(x) = tan hx
Proof: 
Here f(x) = tan hx 
By Maclaurin’s series expansion, 
… (1) 


By equation (1) 
8) then
Proof: 
Here f(x) = log (1 + x) 
By Maclaurin’s series expansion, 
… (1) 





By equation (1) 
9)
In above result we replace x by x
Then
10) Expansion of tan h1x
We know that 
Thus 
11)Expansion of (1 + x)m
Proof: 
Let f(x) = (1 + x)m 
By Maclaurin’s series. 
… (1) 



By equation (1) we get, 
Note that in above expansion if we replace m = 1 then we get, 
Now replace x by x in above we get, 

Expand by, Maclaurin’s theorem
Q1)
S1)
Here f(x) = log (1 + sin x) 
By Maclaurin’s Theorem, 
… (1) 




…….. 
equation (1) becomes, 
Expand by Maclaurin’s theorem,
log sec x
Solution:
Let f(x) = log sec x 
By Maclaurin’s Expansion’s, 
(1) 




By equation (1) 

Prove that
Solution:
Here f(x) = x cosec x 
= 
Now we know that 

Expand upto x6
Solution:
Here 
Now we know that 
… (1) 
… (2) 
Adding (1) and (2) we get 
Show that
Solution:
Here 
Thus 
Taylor’s Series Expansion:
a) The expansion of f(x+h) in ascending power of x is
b) The expansion of f(x+h) in ascending power of h is
c) The expansion of f(x) in ascending powers of (xa) is,
d) Using the above series expansion we get series expansion of f(x+h) or f(x).
Expansion of functions using standard expansions
Expand in power of (x – 3)
Solution:
Let 
Here a = 3 
Now by Taylor’s series expansion, 
… (1) 
equation (1) becomes. 

Using Taylors series method expand
in powers of (x + 2)
Solution:
Here 
a = 2 
By Taylors series, 
… (1) 
Since 
, , ….. 
Thus equation (1) becomes 

Expand in ascending powers of x.
Solution:
Here 
i.e. 
Here h = 2 
By Taylors series, 
… (1) 

equation (1) becomes, 
Thus 

Expand in powers of x using Taylor’s theorem,
Solution:
Here 
i.e. 
Here 
h = 2 
By Taylors series 
… (1) 







By equation (1) 

Exercise
a) Expand in powers of (x – 2)
b) Expand in powers of (x + 2)
c) Expand in powers of (x – 1)
d) Using Taylors series, express in ascending powers of x.
e) Expand in powers of x, using Taylor’s theorem.
Consider
Then limit of f(x) and g(x) both are zero when then
L becomes form.
This form is called indeterminate form. The other indeterminate formal are , , , , OO, etc.
To evaluate limit in this case we use L – Hospital rule
L – Hospital rule for and 
Statement: 
If takes either or 
Indeterminate form, then 
Provided limit is exist 
If again takes either or . 
Then ; limit is exist 
We continue the procedure until the limit is exist. 
Q1) Evaluate
S1)
Let 
… 
By L – Hospital rule, 
Q2) Evaluate
S2)
Let 
… 
By L – Hospital rule 
Q3) Evaluate
S3)
Let 
… 
By L – Hospital rule 
… 
… 
Q4) Find the value of a, b if
S4)
Let 
… 
By L – Hospital rule 
… 
… 
… (1) 
… 
But 
From equation (1) 

Q5) Evaluate
S5)
Let 
… 
… 
(By L – Hospital Rule) 

Q6) Evaluate
S6)
Let 
… 0o form 
Taking log on both sides we get, 
… 
… 
By L – Hospital Rule 
i.e. 

Q7) Evaluate
S7)
Let 
… 
Taking log on both sides, 
… 
By L – Hospital rule, 
i.e. 

Q8) Evaluate
S8)
Let 
… 
Taking log on both sides, we get 
… 
By L – Hospital Rule, 

Exercise
Evaluate the following limits.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
a) Find the value of a, b, c if
b) If is finite then find the value of p and hence the value of the limit.
c) Find the value of a, b if,
d) Find the value of a and b if,
e) Find the value of a and b if,
Reference Books:
1. Advanced Engineering Mathematics by Erwin Kreyszig (Wiley Eastern Ltd.)
2. Advanced Engineering Mathematics by M. D. Greenberg (Pearson Education)
3. Advanced Engineering Mathematics by Peter V. O’Neil (Thomson Learning)
4. Thomas’ Calculus by George B. Thomas, (AddisonWesley, Pearson)
5. Applied Mathematics (Vol. I & Vol. II) by P.N.Wartikar and J.N.Wartikar Vidyarthi Griha Prakashan, Pune.
6. Linear Algebra –An Introduction, Ron Larson, David C. Falvo (Cenage Learning, Indian edition)