MATH 136
Higher Derivatives
dy
= f ′ ( x) denotes the first derivative of y with
dx
respect to x . If we take the derivative of the derivative, then we have the second
derivative of y with respect to x , which is denoted in any of the following ways:
Let y = f (x ) be a function. Then
d2y
(2)
( x) .
2 = f ′′( x) = f
dx
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If we take the derivative again, then we have the third derivative of y with respect
to x , generally denoted by f (3) ( x) . In general, the n th derivative of y with respect to x
is denoted by f (n) (x ) .
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Example 1. Let f ( x) = tan x and g(x ) = cos(x 3 ) . Compute f ′′ (x ) and g ′′ (x ) .
Solution. First, f ′ (x ) =
d(tan x)
= sec 2 x = (sec x) 2 . Then by the Chain Rule,
dx
f ′′ (x ) = 2(sec x )1 ×
d(sec x)
= 2 sec x × sec x tan x = 2 sec2 x tan x .
dx
Next, by the Chain Rule, g ′( x) = − sin( x3 ) × 3x 2 = −3x 2 sin( x 3 ) . Now by the Product
Rule and Chain Rule, we have
g ′′ (x ) = −6x sin( x 3 ) + (−3x 2 ) cos( x3 ) × 3x 2 = −6x sin( x 3 ) − 9 x 4 cos( x 3 ).
It is often important to re-write or simplify f ′ (x ) before computing f ′′ (x ) .
Example 2. Let f ( x) =
x2
4−x
2 . Evaluate f ′′ (x ) .
2x (4 − x 2 ) − x 2 (−2 x )
8x
=
Solution. By the Quotient Rule, we have f ′ (x ) =
, and
2 2
(4 − x )
(4 − x 2 )2
8(4 − x 2 )2 − 8x × 2(4 − x 2 )(−2 x)
f ′′ (x ) =
(4 − x 2 )4
=
8(4 − x 2 ) + 32x 2 32 + 24x 2
=
.
(4 − x 2 )3
(4 − x 2 )3
Higher order derivatives are used mostly in Calculus II and Differential Equations.
In many cases, we need a pattern for the n th derivative as is illustrated in the following
examples.
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Example 3. Let f ( x) = e −4 x . Find the first, second, and third derivatives with respect to
x . Find a general formula for f (n) (x ) .
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Solution. We have f ′ (x ) = (−4)e−4 x , f ′′ (x ) = 16e −4x , and f (3) ( x) = −64e −4 x .
pattern that develops is f (n) (x ) = (−4)n e −4 x .
The
1
. Find the first, second, and third derivatives with respect to x .
x
Find a general formula for f (n) (x ) .
Example 4. Let f ( x) =
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We have f ( x) = x −1 , so that f ′ (x ) = (−1)x −2 , f (2) (x ) = 2 x −3 , and
6
n!
f (3) ( x) = −6 x −4 = − 4 . The pattern that develops is f (n) (x ) = (−1) n n+1 .
x
x
Solution.
Example 5. Let f ( x) = sin x . Find the first, second, third and fourth derivatives with
respect to x . Find a general formula for f (n) (x ) .
Solution. We have f (1) ( x) = cos x , f (2) (x ) = − sin x , f (3) ( x) = − cos x , f (4) (x ) = sin x ,
then
€ the pattern repeats. We can express this pattern as follows:
f
(n)
(−1)k +1 cos x if n = 2k − 1, k ≥ 1
(x ) = 
(−1)k sin x
if n = 2k, k ≥ 1.