Differential Calculus-I - New Age International
Differential Calculus-I - New Age International Differential Calculus-I - New Age International
4 A TEXTBOOK OF ENGINEERING MATHEMATICS–I Case Case ( (vvvvv): ( ): Suppose m is a negative integer. Let us get m = – p, so that p is a positive integer. Then formula (5) becomes, ⎡ ⎤ ⎢ ⎥ ⎣ + ⎦ P n 1 ( 1) pp ( 1) ( p n 1) a D p n ( ax b) ( ax b ) + − ⋅ + … + − ⋅ = + n n n 12 p 1 p( p 1) ( p n 1) a ( 1) 12 p 1 p n ( ax b ) + ⋅ … − + … + − = − ⋅ ⋅ ⋅ … − + n n ( 1) ( p n 1)! a ( p 1)! ( ax b ) + − ⋅ + − = ⋅ − + In particular n ⎛ 1 ⎞ n ( p n 1)! 1 D ⎜ p ⎟ ( 1) ⎝ x ⎠ ( p 1)! p n x + + − = − ⋅ ⋅ − (4) (4) nth derivative of cos (ax + b) Let y = cos( ax + b). Differentiating this successively, we get y y p n dy y1 = =− asin( ax + b) = acos( ax + b+π/ 2) dx 2 dy = dx 2 2 3 dy = dx 3 3 2 =− a sin( ax+ b+π/ 2) 2 = a cos( ax + b + 2π / 2) 3 =− a sin( ax+ b+ 2 π/ 2) 3 = a cos( ax+ b+ 3 π/ 2) ..... ................................ ..... ................................ n d y n y n = = a cos( ax+ b+ nπ/2) n dx Thus, we obtain the formula n ...(13) ...(14) D [cos( ax b)] n + = a cos( ax + b + nπ / 2) n In particular, ...(15) D (cosx) n = cos( x + nπ/ 2) ...(16) (5) (5) nth derivative of sin(ax + b) Let y = sin( ax + b)
DIFFERENTIAL CALCULUS-I 5 Differentiating successively, we get d dy dx 2 dx dx Thus, we have the formula, y 2 = y = a cos( ax + b) = asin( ax + b + π / 2) 1 2 = y = a cos( ax + b + π / 2) = a sin( ax + b + 2π / 2) 2 3 d y 3 3 3 = y3 = a cos( ax + b + 2 π /2) = a sin( ax + b + 3 π/2) dx ............................................................................ ............................................................................ d n y n n = y = a sin( ax + b + nπ / 2) n D [sin( ax b)] n n + = a sin( ax+ b+ nπ /2) In particular, ...(17) D (sinx) n = sin( x + nπ / 2) ...(18) (6) (6) nth derivative of eax sin (bx + c) Let y = sin( bx + c) e ax dy ax ax y 1 = = ae sin( bx + c) + be cos( bx + c) dx For computation of higher-order derivatives it is convenient to express the constants a and b in terms of the constants k and a defined by a = k cosα, b = ksinα So that Thus, Therefore, 2 2 −1 k= a + b , α= tan ( b/ a) dy ax y 1 = = e [ k(cos α )sin( bx + c) + k(sin α )cos( bx + c)] dx 2 ax = ke sin( bx + c +α) d y y ax ax 2 = = k[ ae sin( bx + c +α ) + be cos( bx + c +α)] 2 dx ax = ke [ k(cos α )sin( bx+ c+α ) + k(sin α ) cos( bx+ c+α)] 2 ax = k e sin( bx + c + 2α) 2
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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)
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