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1.1 INTRODUCTION<br />

UNIT I<br />

<strong>Differential</strong> <strong>Calculus</strong>-I<br />

<strong>Calculus</strong> is one of the more beautiful intellectual achievements of human being. The mathematical<br />

study of change motion, growth or decay is calculus. One of the most important idea of differential<br />

calculus is derivative which measures the rate of change of given function. Concept of derivative is<br />

very useful in engineering, science, economics, medicine and computer science.<br />

2<br />

dy d y<br />

The first order derivative of y denoted by , second order derivative denoted by , third<br />

dx<br />

2<br />

dx<br />

3<br />

d y<br />

order derivative by 3 and so on. Thus by differentiating a function y = f (x),<br />

n times, successively,<br />

dx<br />

n<br />

d y<br />

we get the nth order derivative of y denoted by n or D (y)<br />

dx<br />

n or yn (x). Thus, the process of finding<br />

the differential coefficient of a function again and again is called Successive Differentiation.<br />

1.2 nth DERIVATIVES OF SOME STANDARD FUNCTIONS<br />

Below we obtain formulas for the nth order derivatives of some standard functions.<br />

(1) (1) (1) nth derivative of eax Let y = eax . Then by differentiating y successively, we obtain<br />

dy ax<br />

y = = ae ,<br />

1 dx<br />

2<br />

d y 2<br />

y 2 = = ae<br />

2<br />

dx<br />

3<br />

ax<br />

dy 3 ax<br />

y3 = = ae …<br />

3<br />

dx<br />

................................<br />

................................<br />

yn<br />

n<br />

d y<br />

=<br />

= a<br />

n<br />

dx<br />

n<br />

e<br />

ax


2 A TEXTBOOK OF ENGINEERING MATHEMATICS–I<br />

Thus, we have the formula<br />

In particular,<br />

( )<br />

ax n D e<br />

( )<br />

x n D e<br />

n ax<br />

= a e<br />

...(1)<br />

x<br />

= e<br />

...(2)<br />

(2) (2) nth derivative of log (ax + b)<br />

Let y = log( ax + b).<br />

Then we find, by successive differentiation<br />

y 1<br />

dy a<br />

= =<br />

dx ax + b<br />

2<br />

2<br />

d y a<br />

y2 = = ( −1)<br />

2<br />

2<br />

dx<br />

( ax + b)<br />

3<br />

3<br />

d y<br />

a<br />

y3 = = ( −1)<br />

( −2)<br />

⋅<br />

3<br />

3<br />

dx<br />

( ax + b)<br />

4 4<br />

d y a<br />

y4 = = ( −1)( −2)( −3)<br />

4 4<br />

dx ( ax + b)<br />

... ............................................<br />

... ............................................<br />

yn<br />

Thus, we have the formula<br />

In particular,<br />

D n<br />

n<br />

d y<br />

= = ( −1)<br />

n<br />

dx<br />

n−1<br />

n−1n ( −1) ⋅( n−1)! a<br />

[log( ax + b)]<br />

=<br />

n<br />

( ax + b)<br />

D [logx]<br />

n<br />

(3) (3) nth derivative of (ax + b) m<br />

n−1<br />

( −1) ⋅( n −1)!<br />

=<br />

n<br />

x<br />

Let y m<br />

= ( ax + b)<br />

Differentiating successively, we get<br />

y1<br />

y2<br />

=<br />

dy<br />

= ma ( ax + b)<br />

dx<br />

2<br />

( n −1)!<br />

a<br />

⋅<br />

( ax + b)<br />

m−1<br />

d<br />

y<br />

2<br />

= = m(<br />

m −1)<br />

⋅ a ( ax + b)<br />

2<br />

dx<br />

n<br />

n<br />

m−2<br />

...(3)<br />

...(4)


DIFFERENTIAL CALCULUS-I 3<br />

3<br />

d y<br />

3 m−2<br />

y3<br />

= = m(<br />

m −1)<br />

( m − 2)<br />

a ( ax + b)<br />

3<br />

dx<br />

........................................................<br />

........................................................<br />

yn<br />

n m−n<br />

( m −1)<br />

( m − 2)<br />

L ( m − n + 1)<br />

a ( ax + b<br />

...(5)<br />

= m<br />

)<br />

This formula is true for all m.<br />

Following are some particular cases.<br />

Case Case ( (iiiii): ( ( ): Suppose m = n (a +ve integer)<br />

In equation (5) becomes,<br />

n<br />

n<br />

n n n<br />

D [( ax + b)<br />

] = nn ( 1) ( n 2) 1 a( ax b) −<br />

− − ⋯ ⋅ +<br />

n<br />

= n! a<br />

...(6)<br />

In particular,<br />

( )<br />

n n D x = n!<br />

...(7)<br />

Case Case (<br />

( (ii ii): ii ii ii ): Suppose m is a positive integer and m > n. Then formula (5) becomes<br />

n<br />

D [( ax + b)<br />

m<br />

]<br />

m( m−1) …( m− n+ 1)( m−n)( m−n−1) …2⋅1<br />

= ⋅ a ( ax+ b)<br />

( m−n)( m−n−1) …2⋅1<br />

m! n m − n<br />

= a ( ax + b)<br />

( m − n)<br />

!<br />

In particular,<br />

( )<br />

m n m! m − n<br />

D x = x<br />

( m − n)<br />

!<br />

Case Case ( (<br />

( (iii iii): iii iii iii ): Suppose m is a positive integer and n > m. From (6) we note that<br />

n m−n ...(8)<br />

...(9)<br />

n m<br />

D [( ax + b)<br />

] ! m<br />

= m a<br />

In differentiate further, the right-hand side gives zero.<br />

Thus, n m<br />

D [( ax + b)<br />

] = 0 if n > m ...(10)<br />

In particular,<br />

( )<br />

m n D x = 0 for n > m ...(11)<br />

Case Case ( (<br />

( (iv iv): iv iv iv ): Suppose m = – 1, in this case formula (5) becomes,<br />

n ⎡ 1 ⎤<br />

n<br />

D ⎢ax + b⎥<br />

= ( −1)<br />

( −2)<br />

( −3)<br />

K(<br />

−n)<br />

a ( ax + b)<br />

⎣ ⎦<br />

n n<br />

( 1) n! a<br />

( ax b )<br />

+<br />

− ⋅<br />

=<br />

+<br />

n 1<br />

−1−n<br />

...(12)


4 A TEXTBOOK OF ENGINEERING MATHEMATICS–I<br />

Case Case ( (vvvvv): ( ): Suppose m is a negative integer. Let us get m = – p, so that p is a positive integer. Then<br />

formula (5) becomes,<br />

⎡ ⎤<br />

⎢ ⎥<br />

⎣ + ⎦<br />

P<br />

n 1 ( 1) pp ( 1) ( p n 1) a<br />

D<br />

p n<br />

( ax b)<br />

( ax b )<br />

+<br />

− ⋅ + … + − ⋅<br />

=<br />

+<br />

n n<br />

n 12 p 1 p( p 1) ( p n 1) a<br />

( 1)<br />

12 p 1 p n<br />

( ax b )<br />

+<br />

⋅ … − + … + −<br />

= − ⋅ ⋅<br />

⋅ … − +<br />

n n<br />

( 1) ( p n 1)! a<br />

( p 1)! ( ax b )<br />

+<br />

− ⋅ + −<br />

= ⋅<br />

− +<br />

In particular<br />

n ⎛ 1 ⎞ n ( p n 1)! 1<br />

D ⎜ p ⎟ ( 1)<br />

⎝ x ⎠ ( p 1)! p n<br />

x +<br />

+ −<br />

= − ⋅ ⋅<br />

−<br />

(4) (4) nth derivative of cos (ax + b)<br />

Let y = cos( ax + b).<br />

Differentiating this successively, we get<br />

y<br />

y<br />

p n<br />

dy<br />

y1<br />

= =− asin( ax + b) = acos( ax + b+π/<br />

2)<br />

dx<br />

2<br />

dy<br />

=<br />

dx<br />

2 2<br />

3<br />

dy<br />

=<br />

dx<br />

3 3<br />

2<br />

=− a sin( ax+ b+π/<br />

2)<br />

2<br />

= a cos( ax + b + 2π<br />

/ 2)<br />

3<br />

=− a sin( ax+ b+<br />

2 π/<br />

2)<br />

3<br />

= a cos( ax+ b+<br />

3 π/<br />

2)<br />

..... ................................<br />

..... ................................<br />

n<br />

d y n<br />

y n = = a cos( ax+ b+ nπ/2)<br />

n<br />

dx<br />

Thus, we obtain the formula<br />

n<br />

...(13)<br />

...(14)<br />

D [cos( ax b)]<br />

n<br />

+ = a cos( ax + b + nπ<br />

/ 2)<br />

n In particular,<br />

...(15)<br />

D (cosx)<br />

n = cos( x + nπ/<br />

2)<br />

...(16)<br />

(5) (5) nth derivative of sin(ax + b)<br />

Let y =<br />

sin( ax + b)


DIFFERENTIAL CALCULUS-I 5<br />

Differentiating successively, we get<br />

d<br />

dy<br />

dx<br />

2<br />

dx<br />

dx<br />

Thus, we have the formula,<br />

y<br />

2<br />

= y = a cos( ax + b)<br />

= asin(<br />

ax + b + π / 2)<br />

1<br />

2<br />

= y = a cos( ax + b + π / 2)<br />

= a sin( ax + b + 2π<br />

/ 2)<br />

2<br />

3<br />

d y<br />

3<br />

3 3<br />

= y3 = a cos( ax + b + 2 π /2) = a sin( ax + b + 3 π/2)<br />

dx<br />

............................................................................<br />

............................................................................<br />

d<br />

n<br />

y<br />

n<br />

n<br />

= y = a sin( ax + b + nπ<br />

/ 2)<br />

n<br />

D [sin( ax b)]<br />

n<br />

n + = a sin( ax+ b+ nπ<br />

/2)<br />

In particular,<br />

...(17)<br />

D (sinx)<br />

n = sin( x + nπ<br />

/ 2)<br />

...(18)<br />

(6) (6) nth derivative of eax sin (bx + c)<br />

Let y = sin( bx + c)<br />

e ax<br />

dy ax<br />

ax<br />

y 1 = = ae sin( bx + c)<br />

+ be cos( bx + c)<br />

dx<br />

For computation of higher-order derivatives it is convenient to express the constants a and b in<br />

terms of the constants k and a defined by<br />

a = k cosα, b = ksinα<br />

So that<br />

Thus,<br />

Therefore,<br />

2 2 −1<br />

k= a + b , α= tan ( b/ a)<br />

dy ax<br />

y 1 = = e [ k(cos α )sin( bx + c) + k(sin α )cos( bx + c)]<br />

dx<br />

2<br />

ax<br />

= ke sin( bx + c +α)<br />

d y<br />

y ax ax<br />

2 = = k[ ae sin( bx + c +α ) + be cos( bx + c +α)]<br />

2<br />

dx<br />

ax<br />

= ke [ k(cos α )sin( bx+ c+α ) + k(sin α ) cos( bx+ c+α)]<br />

2 ax<br />

= k e sin( bx + c + 2α)<br />

2


6 A TEXTBOOK OF ENGINEERING MATHEMATICS–I<br />

Proceeding like this, we obtain<br />

dx<br />

Thus, we have the formula<br />

D ax n<br />

n<br />

d y<br />

y n = n ax<br />

n = k e sin( bx + c + nα)<br />

2 2 n /2 −1<br />

= ( a + b ) sin{ bx+ c+ ntan ( ba / )}<br />

...(19)<br />

[ e sin( bx + c)]<br />

In particular,<br />

n x<br />

D [ e sin x ] 2 sin( / 4)<br />

2 /<br />

= ⋅ e x + nπ<br />

x n ...(20)<br />

(7) (7) nth derivative of eax cos(bx + c)<br />

Let y = cos( bx + c)<br />

Then<br />

Therefore,<br />

dy<br />

y 1 =<br />

dx<br />

2<br />

e ax<br />

= ae<br />

ax<br />

ax<br />

ax<br />

cos( bx + c)<br />

− be sin( bx + c)<br />

= e [ k(cos α )cos( bx + c) −k(sin α )sin( bx + c)]<br />

= cos( bx + c + α)<br />

ke ax<br />

d y<br />

y 2 = ax<br />

ax<br />

= k[<br />

ae cos( bx + c + α)<br />

− be sin( bx + c + α)]<br />

2<br />

dx<br />

= k e cos( bx + c + 2α)<br />

Proceeding like this, we obtain<br />

n<br />

d y<br />

y n = n ax<br />

= k e cos( bx + c + nα)<br />

n<br />

dx<br />

Thus, we have the formula<br />

n ax<br />

D [ e cos( bx+ c)]<br />

In particular,<br />

ax<br />

= ke [ k(cos α ) cos( bx+ c+α) −k(sin α )sin( bx+ c+α)]<br />

2<br />

ax<br />

D x n n/2 x<br />

2 2 n/2 ax<br />

−1<br />

= ( a + b ) e cos{ bx+ c+ ntan ( ba / )}<br />

...(21)<br />

( e cosx)<br />

= 2 e cos( x+ nπ<br />

/4)<br />

...(22)<br />

(8) (8) nth derivative of amx Let y mx<br />

= a<br />

Taking logarithm on both sides,<br />

logy = mx loga<br />

Differentiating w.r.t. ‘x’, we get<br />

1 dy<br />

⋅ = m<br />

loga ⋅1<br />

y dx


DIFFERENTIAL CALCULUS-I 7<br />

dy<br />

y 1 =<br />

dx<br />

2<br />

= ( m loga)<br />

⋅ y<br />

d y<br />

y 2 = = m loga ⋅ y<br />

2<br />

1<br />

dx<br />

= ( m loga)(<br />

m loga)<br />

⋅ y<br />

3<br />

= ( mloga)<br />

d y<br />

2<br />

y 3 = = ( mlog a) ⋅y<br />

3<br />

dx<br />

= ( mloga)<br />

2<br />

3<br />

y<br />

y<br />

........................<br />

........................<br />

n<br />

d y<br />

y n = n<br />

m a y<br />

n = ( log ) ⋅<br />

dx<br />

n mx<br />

[ ]<br />

D a = ( mlog a) ⋅a<br />

1<br />

n mx<br />

WORKED EXAMPLES<br />

Example Example 1: 1: Find the nth derivative of the following functions:<br />

Solution:<br />

Solution:<br />

(i) 3<br />

y = sin x<br />

(i) sin 3 x (ii) cos 4 x.<br />

where 3<br />

x<br />

\ (sin )<br />

3 D x<br />

n<br />

(ii) y = cos 4 x<br />

where 2<br />

x<br />

so that 4<br />

cos x<br />

1<br />

sin = ( 3sin<br />

x − sin3x)<br />

4<br />

3 n 1 n<br />

= D (sinx)<br />

− D (sin3x)<br />

4 4<br />

=<br />

3 ⎛ nπ<br />

⎞ 3 ⎛ nπ<br />

⎞<br />

sin⎜<br />

x + ⎟ − sin⎜3x<br />

+ ⎟<br />

4 ⎝ 2 ⎠ 4 ⎝ 2 ⎠<br />

1<br />

cos = ( 1 + cos2x)<br />

2<br />

1<br />

= ( 1+<br />

cos2x)<br />

4<br />

1 2<br />

=<br />

[ 1+<br />

cos 2x<br />

+ 2cos2x]<br />

4<br />

2<br />

n<br />

[Using the formula (15)]


8 A TEXTBOOK OF ENGINEERING MATHEMATICS–I<br />

where cos 2x<br />

2<br />

\ 4<br />

x<br />

\<br />

1<br />

= ( 1 + cos4x)<br />

2<br />

1 ⎡ 1<br />

⎤<br />

cos = ⎢1<br />

+ ( 1+<br />

cos4x)<br />

+ 2 cos2x<br />

4<br />

⎥<br />

⎣ 2<br />

⎦<br />

=<br />

1<br />

4<br />

+<br />

1<br />

8<br />

1 1<br />

+ cos4x<br />

+ cos2x<br />

8 2<br />

4 3 1 1<br />

cos x = + cos2x<br />

+ cos4x<br />

8 2 8<br />

n 4 n ⎛ 3⎞<br />

1 n 1 n<br />

D (cos x ) = D ⎜ ⎟ + D (cos2x)<br />

+ D (cos4x)<br />

⎝ 8 ⎠ 2<br />

8<br />

n n<br />

2 ⎛ nπ⎞ 4 ⎛ nπ⎞<br />

= 0 + cos 2x cos 4x<br />

2<br />

⎜ +<br />

2<br />

⎟+ +<br />

8<br />

⎜<br />

2<br />

⎟<br />

⎝ ⎠ ⎝ ⎠<br />

n n<br />

2 ⎛ nπ⎞ 4 ⎛ nπ⎞<br />

= cos 2x cos 4x<br />

2<br />

⎜ +<br />

2<br />

⎟+ +<br />

8<br />

⎜<br />

2<br />

⎟<br />

⎝ ⎠ ⎝ ⎠<br />

Example Example 2: 2: Find the nth derivative of the following:<br />

(i) sin h 2x<br />

sin4x<br />

(ii)<br />

−x<br />

2<br />

e sin x<br />

(iii) e x<br />

(v)<br />

Solution:<br />

Solution:<br />

x<br />

(i)<br />

1 2x<br />

−2x<br />

sinh<br />

2x<br />

sin4x<br />

= ( e − e ) ⋅ sin4x<br />

2<br />

sin h2x x 3 2 cos (iv) e x x<br />

x<br />

e x<br />

−<br />

sin cos 2<br />

ax<br />

cos sin<br />

2<br />

1 2x<br />

−2x<br />

sin h2x sin4x<br />

= [ e sin4x<br />

− e sin4x]<br />

2<br />

n 1 n 2x n −2x<br />

D (sinh 2x<br />

sin4x)<br />

= [ D ( e sin4 x) −D<br />

( e sin4 x)]<br />

2<br />

2x −<br />

−2x<br />

e e<br />

=<br />

2<br />

1 n/2 2x −1−2x −1<br />

[20 { e sin(4x ntan2) e sin(4x ntan2)}]<br />

=<br />

2<br />

+ − −<br />

Using the formula (19)


DIFFERENTIAL CALCULUS-I 9<br />

1<br />

(ii) We have sin ( 1 cos2<br />

)<br />

2<br />

2<br />

x = − x<br />

Therefore,<br />

n −x 2 1 n −x 1 n −x<br />

x = −<br />

D ( e<br />

sin<br />

1<br />

(iii) We have cos (cos3<br />

3cos<br />

)<br />

4<br />

3<br />

x = x + x<br />

n<br />

D ( e<br />

2x<br />

3<br />

)<br />

D ( e ) D ( e cos2 x)<br />

2 2<br />

1 −x n 1 n/2 −x −1<br />

= e ( −1) − [5 e cos(2x+ ntan<br />

( −2))]<br />

2 2<br />

1 −x n n/2<br />

−1<br />

= e [( −1) −5cos(2x−ntan (2))]<br />

2<br />

cos x)<br />

= D ( e<br />

4<br />

1 n 2x<br />

2x<br />

3 n<br />

cos3x)<br />

+ D ( e<br />

4<br />

cosx)<br />

1 2 2 n/2 2x −1<br />

[(2 3 ) e cos{3x ntan<br />

(3/ 2)}]<br />

= + +<br />

4<br />

3 2 2 / 2 2<br />

−1<br />

e x n<br />

+ [( 2 + 1 ) cos{ x + n tan<br />

4<br />

1 2x n / 2<br />

−1<br />

= e [ 13 cos{ 3x<br />

+ n tan<br />

4<br />

n / 2<br />

−1<br />

+ 3(<br />

5)<br />

⋅ cos{ x + n tan<br />

1<br />

(iv) We have sin x cos2x<br />

= (sin3x<br />

− sin x)<br />

2<br />

Therefore,<br />

n −x<br />

1<br />

D ( e sin x cos2x)<br />

= D ( e<br />

2<br />

(v) We note that<br />

= [(( −1)<br />

2<br />

1 n<br />

sin3x)<br />

− D ( e<br />

2<br />

n −x<br />

−x<br />

+ 3<br />

)<br />

( 3/<br />

2)}<br />

( 1/<br />

2)}]<br />

sin x)<br />

1 2 2 / 2 −<br />

−1<br />

e x n<br />

− [(( −1)<br />

2<br />

= e<br />

2<br />

sin{ 3x<br />

+ n tan<br />

1 2 2 / 2 −<br />

−1<br />

+ 1 ) e sin{ x + n tan<br />

x n<br />

1 −x<br />

n / 2<br />

−1<br />

n / 2<br />

[ 10<br />

sin( 3x<br />

− ntan<br />

2 1<br />

cos xsinx =<br />

( 1 + cos2x)<br />

⋅ sin x<br />

2<br />

3)<br />

− 2<br />

( 1/<br />

2)}]<br />

( −3)}]<br />

( −1)}]<br />

sin( x − nπ<br />

/ 4)]


10 A TEXTBOOK OF ENGINEERING MATHEMATICS–I<br />

\ 2<br />

D ( e cos xsin<br />

x)<br />

ax n<br />

=<br />

1 1<br />

= sin x + ⋅ 2sin<br />

x cos2x<br />

2 4<br />

1 1<br />

= sin x + (sin3x<br />

− sin x)<br />

2 4<br />

1 1<br />

= sin x + sin3x<br />

4 4<br />

1<br />

4<br />

n<br />

ax<br />

D ( e<br />

1 n<br />

sinx)<br />

+ D ( e<br />

4<br />

ax<br />

sin3x)<br />

1 ax 2 n /2 −1<br />

= e [( a + 1) sin{ x+ ntan (1/ a)}<br />

4<br />

n / 2<br />

−1<br />

+ ( a + 9)<br />

sin{ 3x<br />

+ n tan ( 3/<br />

a)}]<br />

Example Example 3: 3: 3: Find the nth derivative of the following:<br />

(i)<br />

Solution:<br />

Solution:<br />

x + 3<br />

( x −1)(<br />

x + 2)<br />

(i)<br />

x + 3<br />

y =<br />

( x −1)(<br />

x + 2)<br />

By partial fractions<br />

2<br />

(ii)<br />

x + 3<br />

( x −1)(<br />

x + 2)<br />

A B<br />

= +<br />

x − 1 x + 2<br />

Then x + 3 = Ax ( + 2) + Bx ( −1)<br />

Taking x = 1 ⇒ A = 4/3<br />

Equation (1), becomes<br />

x =−2 ⇒ B =−1/3<br />

x + 3 4 1 1 1<br />

= ⋅ −<br />

( x −1)(<br />

x + 2)<br />

3 ( x− 1) 3( x+<br />

2)<br />

2<br />

x<br />

( x + 2)(<br />

2x<br />

+ 3)<br />

⎡ x + 3 ⎤<br />

D ⎢<br />

⎥<br />

⎣(<br />

x −1)(<br />

x + 2)<br />

⎦<br />

n 4 n⎛ 1 ⎞ 1 n⎛<br />

1 ⎞<br />

= D D<br />

3<br />

⎜ −<br />

x− 1<br />

⎟<br />

3<br />

⎜<br />

x+<br />

2<br />

⎟<br />

⎝ ⎠ ⎝ ⎠<br />

n n<br />

4( −1) ⋅n! 1( −1) ⋅n!<br />

= −<br />

3 ( x − 1) 3(<br />

x + 2)<br />

n+ 1 n+<br />

1<br />

1 n ⎡ 4 1 ⎤<br />

= ( −1) ⋅n! ⎢ −<br />

3 n+ 1 n+<br />

1⎥<br />

⎣( x− 1) ( x+<br />

2) ⎦<br />

...(1)<br />

[Using the formula (12)]

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