Chapter 3 | Derivatives
241
Higher-Order Derivatives The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as higher-order derivatives . The notation for the higher-order derivatives of y = f ( x ) can be expressed in any of the following forms:
f ″( x ), f ‴( x ), f (4) ( x ),…, f ( n ) ( x ) y ″( x ), y ‴( x ), y (4) ( x ),…, y ( n ) ( x )
d 3 y dx 3
d 2 y dx 2
d 4 y dx 4
d n y dx n .
,
,
,…,
may be viewed as an attempt to express d dx ⎛ ⎝ dy dx ⎞
d 2 y dx 2
⎠ more compactly.
It is interesting to note that the notation for
⎛ ⎝ ⎜ d 2 y dx 2
⎞ ⎠ ⎟ =
Analogously, d dx ⎛
⎛ ⎝ dy dx
⎞ ⎠
⎞ ⎠ = d
d 3 y dx 3
⎝ d
.
dx
dx
Example 3.15 Finding a Second Derivative
For f ( x ) =2 x 2 −3 x +1, find f ″( x ).
Solution First find f ′( x ).
Substitute f ( x ) =2 x 2 −3 x +1 and f ( x + h ) =2( x + h ) 2 −3( x + h )+1 into f ′( x ) = lim h →0 f ( x + h )− f ( x ) h .
f ′( x ) = lim h →0 ⎛
⎝ 2( x + h ) 2 −3( x + h )+1 ⎞
⎠ −(2 x 2 −3 x +1)
h
4 xh +2 h 2 −3 h h
= lim
Simplify the numerator.
h →0
Factor out the h in the numerator and cancel with the h in the denominator.
(4 x +2 h −3)
= lim
h →0
=4 x −3
Take the limit.
Next, find f ″( x ) by taking the derivative of f ′( x ) =4 x −3.
f ( x + h )− f ( x ) h
Use f ′( x ) = lim h →0 place of f ( x ).
with
f ′( x ) in
f ′( x + h )− f ′( x ) h
f ″( x ) = lim h →0
Substitute f ′( x + h ) =4( x + h )−3and f ′( x ) =4 x −3.
⎛ ⎝ 4( x + h )−3 ⎞
⎠ −(4 x −3)
= lim = lim
h
h →0
4
Simplify.
h →0
=4
Take the limit.
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