
9-16 Calculus – Differentiation and Integration
Differentiating Eq. (9.1) yields
f ' x a a x a x n a x na x
n
n
n
n
( ) ( ) ,= + + + + − +
−
− −
1 2 3
2
1
2 1
2 3 1
and now f ′(0) = a
1
.
Again differentiating f ′(x), we obtain
f '' x a a x n n a x n n a x
n
n
n
n
( ) ( )( ) ( ) ,= + ⋅ + + − − + −
−
− −
2 3 2 1 2 1
2 3 1
3 2
and now f ′′(0) = 2a
2
.
Again differentiating f ′′(x), we obtain f ′′′(0) = 3!a
3
.
And after k differentiations, we get
f k a k n
k
k
( )
( ) ! , .0 = ≤provided
So, we can rewrite the polynomial f(x), using its value and the value of its derivatives at 0, as follows:
f x f f ' x
f ''
x
f '''
x
f
n
x
n
( ) ( ) ( )
( )
!
( )
!
( )
( )!
( )
= + + + + +
−
−
0 0
0
2
0
3
0
1
2 3
1