Module 2

Differential Calculus- I

2.1 Introduction to limits, continuity, and differentiability

Limits

1). Compute limx→−2(3x2+5x−9)

Solution:

First, use property 2 to divide the limit into three separate limits. Then use property 1 to bring the constants out of the first two. This gives,

Lim x→−2(3x2+5x−9)=lim x→−2(3x2)+lim x→−2(5x)−lim x→−2(9)

=3(−2)2+5(−2)−(9)

= -7

2). Find the limits of limx→3 [x(x+2)].

Solution: limx→3 [x(x+2)] = 3(3+2) = 3 x 5 = 15

Continuous Function

A function f(x) is said to be continuous at x = a if

i)       f(a) exist is function must exist at x = a

ii)    

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 the converse need not be true.

2.2 Rolle’s Theorem

If

i)      f(x) is continuous in the closed [a, b]

ii)    f(x) is differentiable in (a, b) &

iii)  f(a) = f(b)

Then there exist at least one value ‘c’ in (a, b) such that f’(c) = 0.

Exercise 1

Verify Rolle’s theorem for the function f(x) = x2 for

Solution:

Here f(x) = x2;

i)      Since f(x) is algebraic polynomial which is continuous in [-1, 1]

ii)    Consider f(x) = x2

Diff. w.r.t. x we get

f'(x) = 2x

Clearly f’(x) exists in (-1, 1) and does not becomes infinite.

iii)  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.

Exercise 2

Verify Rolle’s Theorem for the function f(x) = ex(sin x – cos x) in

Solution:

Here f(x) = ex(sin x – cos x);

i)      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 .

ii)    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 .

iii)  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.

Exercise 3

Verify whether Rolle’s theorem is applicable or not for

Solution:

Here f(x) = x2;

i)      X2 is an algebraic polynomial hence it is continuous in [2, 3]

ii)    Consider

F’(x) exists for each

iii)  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]

Exercise 4

Show that between any two real roots of an equation

, is at least one real root of

.

Lagrange’s Mean value Theorem:-

Statement:- If

i)      f(x) is continuous in [a, b]

ii)    f(x) is differentiable in (a, b) then there exist at least one value such that

Exercise 5

Verify the Lagrange’s mean value theorem for

Solution:

Here

i)      Clearly f(x) = log x is logarithmic function. Hence it is continuous in [1, e]

ii)    Consider f(x) = log x.

Diff. w.r.t. x we get,

Clearly f’(x) exists 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.

Exercise 6

Verify mean value theorem for f(x) = tan-1x in [0, 1]

Solution:

Here ;

i)      Clearly is an inverse trigonometric function and hence it is continuous in [0, 1]

ii)    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 decreasingfunction.

Exercise 7

Prove that

And hence show that

Solution:

Let ; 

i)      Clearly is a logarithmic function and hence it is continuous also

ii)    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

Exercise 8

Prove that , use mean value theorem to prove that,

Hence show that

Solution:

i)     Let f(x) = sin-1x; 

ii)    Clearly f(x) is inverse trigonometric function and hence it is continuous in [a, b]

iii)  Consider f(x) = sin-1x

diff. w.r.t. x we get,

Clearly f’(x) is finite and exists 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

2.3 Lagrange’s Mean value theorem and Cauchy mean value theorem

Lagrange’s Mean Value Theorem:

Statement:

Let f: [a,b] be a continuous function ,differentiable on the open interval (a,b).Thenthere exista some c (a,b) such that

f’(c) =

proof:

we reduce the problem to Rolle’s theorem by using auxillary functions.

Consider,

        g(x) = f(x)- 

note: g(a)=g(b)=f(b)

by rolle’s theorem therexists, c in (a,b) such that g’(c) =0 or s

  f’(c)  -  = 0

Which simplifies to,

f’(c)  =  

exercise 9:

use lagranges mean value theorem to determine a point P on the curve y=

where the tangent is parallel to the chord joining (2,0) and (3,1)

solution:

consider y= in [2,3]

(i)                Function is continuos in[2,3] as  algebraic expression with positive exponent is continuous.

(ii)              y’= ,  y’ exists in (2,3) hence the function is derivable in (2,3)

hence the condition of LMV theorem is satisfied.

Hence, there exists one c in (2,3) such that =

  = 4(c-2) = 1 4c=9  c= 4/9

for x = 9/4, tangent is parallel to the chord joining (2,0) and (3,1)

Substituting in (i) we get,

Y=   = = ½

Exercise 10:

Verify lagrange’s mean value theorem for the following function

f(x) =

putting x=a=2 and x=b=5 ,we get

f(2) =

f(5) =

clearly,

f(2) f(5)

since f(x) is a polynomial function in x, then f(x) is continuous in [2,5].

And f(x) is polynomial in x,then it can be differentiatie such that f’(x) = 4x-7

Then by LMV theorem there exists c (2,5) such that’

 f’ (c) =

4c-7 =

c=3.75

Hence lagrange’s mean value theorem is verified for f(x) in [2,5].

Cauchy’s Mean Value Theorem:

Statement:-

If f(x) and g(x) are any two functions such that

a)     f(x) and g(x) are continuous in (a, b)

b)    both f(x) and g(x) are derivable in (a, b)

c)    

Then for any value of , at least such that

Exercise 11

Verify Cauchy mean value theorems for &in

Solution:

Let &;

i)      Clearly f(x) and g(x) both are trigonometric functions. Hence continuous in

ii)    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

iii)   

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.

Exercise 12

Considering the functions ex and e-x, show that c is arithmetic mean of a & b.

Solution:

i)      Clearly f(x) and g(x) are exponential functions Hence they are continuous in [a, b].

ii)    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

Exercise 13

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:, ]

2.4 Successive Differentiation (nth order derivatives)

Exercise 14:

Find the nth derivative of sin3 x

  Sol: we know that sin 3x= 3sin x 4sin3 x =  sin3x  =    

Differentiate   n    times w.r.t x,

( sin3 x) =   (3 sinx- sin3x)

                  =   (  -3n. Sin(  3x+ nπ/2)  +   3 sin (x+  nπ/2)) nϵz

  Exercise 15:

Find the nth derivative of sin 5x. sin 3x.?

Sol: let y =   sin 5x.sin 3x= ( sin 5x.sin 3x)  

            ⇒y= (  cos 2x -  cos8x)

           ⇒ y= ( cos 2x- cos8x )

 Differentiate n times w.r.t x,

 Yn  = ( cos 2x -  cos8x )

⇒ yn = ( 2 n (cos( 2x+ nπ/2)-  8n .cos (8x + nπ/2)) nϵz.

Successive n th derivative of nth elementary function ie., exponential

Exercise 16:

If  y  = ae n x +  be –nx ,    then show that   y2= n2y

 Sol: Y= aenx + be-nx

  y 1 =   a.n.enx  - b.n.e-nx

 y2 = an2 enx – bn2 e-nx  = n2 (ae nx+ be –nx)

 y2= n2y.

 Exercise 17:

If  y= e-kx/2(a cosnx+ b sinnx) then show that.,y2+ ky1+(n2+ k2/4)y =0

Sol : y= e-kx/2(a cosnx+ b sinnx)

Differentiating  w.r.to. x.,

Y1 = e-kx/2( -an sin nx + bn cos nx) - k/2.y

Y1+ k/2.y = ne-kx/2 ( -an sin nx + bn cos nx)    (1)

Differentiating w.r.to x.,

Y2+ k/2.y1 = ne-kx/2  (-k/2) ( -an sin nx + bn cos nx) + n e-kx/2(-an  cosnx- bn  sinnx).

= -(k/2) (y1+ k/2 y)- n2 y = - (k/2 y1)- ( k2/4)y- n2y.

  y2 + ky1 +(n2+ k2/4)y = 0.

Successive differentiation of nth derivative of elementary functions ie., logarithmic

Example 18:

If  Y 2)  =  log( x + 2) ) then show that (1 + x 2) y1  +xy =1

Sol:  Y 2)  =  log( x + 2) )

Differentiating w.r.to x.,

y.1/ 2  2 .2x+ 2. Y1.

 =(1+x ) y +xy = . 1/x 2  . 2)  +x/ 2 =1

2.5 Leibnitz theorem and its application

Statement:

If F(x)  = ,then

Proof:

Let P(t) =

F(x) =

F(x) =

= P’(h(x))h’(x) – P’(g(x))g’(x)

         = f(h(x)).h’(x)-f(g(x))g’(x)

Exercise 19:

Sove the following using lebinitz rule:

Solution:

=0

2.6 Envelope, Involutes and Evolutes

Exercise 20:

Find the envelope of the family of the ellipses defined by the equation

Solution:

The equation for the given family of curves can be writeen as

=1…..(1)

Where the semi axis a is the parameter and 0<a<1

Differentiating eq(1) w.r.to a  is

Take the square root of both sides of the equation:

Express from the last equation

Substituting in the eq(1) we get

=1

=

Exercise 21:

Find the evolute for the following parabola  y=

Solution:

For evoluting the parabola we have the following formulae,

Substituting the given function we get

By eliminating the x in we getx =

Substituting in we get

Therefore the required answer is

2.7 Curve tracing: Cartesian and Polar co-ordinates

Exercise 22:

For a given curve  convert the point (2,) from polar to Cartesian form

Solution:

  r=2 and =   we have,

x = rcos and y= rsin

x= 2.cos = 2.1/2 = 1

y = 2.sin = 2./2 =

Therefore the required Cartesian point is (1,).

Reference Books-

1.E. Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, 2005.

2.Peter V. O’Neil, Advanced Engineering Mathematics, Thomson (Cengage) Learning, 2007.

3.Maurice D. Weir, Joel Hass, Frank R. Giordano, Thomas, Calculus, Eleventh Edition,Pearson.

4.D. Poole, Linear Algebra: A Modern Introduction, 2nd Edition, Brooks/Cole, 2005.

5.Veerarajan T., Engineering Mathematics for the first year, Tata McGraw-Hill, New Delhi, 2008.

6.Ray Wylie C and Louis C Barret, Advanced Engineering Mathematics, Tata Mc-Graw-Hill, Sixth Edition.

7.P. Sivaramakrishna Das and C. Vijayakumari, Engineering Mathematics, 1st Edition, PearsonIndia Education Services Pvt. Ltd

8. Advanced Engineering Mathematics. Chandrika Prasad, Reena Garg, 2018.

9. Engineering Mathematics – I. Reena Garg, 2018.