Differential Calculus-I - New Age International

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