Higher Order Derivatives
If a function is differentiable, then we can find its
derivative. If this derivative is differentiable then we can find
the derivative of this derivative. The last derivative is the
second derivative of the first function. At each step
of the way, if the new function is differentiable, we can continue
taking the derivative of each derivative in turn.
The notation we use for higher order derivatives
is:
first derivative

f '(x)


second derivative

f ''(x)


third derivative

f '''(x)


fourth derivative

f^{(4)}(x)


fifth derivative

f^{(5)}(x)


n^{th} derivative

f^{(n)}(x)


The primed notation is used for the first three, but
after three it becomes too unwieldy. So we then use numbers within parenthesis.
f^{(4)}(x) is the fourth derivative of f(x).
The parenthesis around the 4 is required; otherwise we'd have
f^{4}(x) and f^{4}(x) is f(x) raised to the
4^{th} power.
Recall that is an operator, not a fraction. It means
“the derivative with respect to x.” should be interpreted as so and we find the derivatives of y, one after the
other, from the inside out.
Example, suppose y = x^{1} then for x
≠ 0,
and working from the inside we have:
and next:
finally:
Another way of looking at this example is to take the
derivative of x^{1} three times one after the other. The
following examples take this approach.
When working with derivatives of higher order take your time and study the
problem at hand. If fractions are involved consider rewriting them as products
of factors with negative exponents first. Then again, using the reciprocal rule for
derivatives is better suited. If you can simplify the expression first then do so.
Above all, take your time and write every step along the way to the solution.
Example:

find f''(x) for f(x) = ½ x^{2} + x + 1



the first derivative:
f'(x) = 2( ½ x^{21} )+
1x^{11} + 0
f'(x) = x^{1} + x^{0}
f'(x) = x + 1
now the second derivative:
f''(x) = 1(x^{11}) + 0
f''(x) = x^{0} = 1

Every step is shown, recall that the derivative of the
derivative of the sum of the functions is the sum of the
derivatives of the functions.




Example:

find f^{(5)}(x) for f(x) = x^{5} +
C^{2}



the first derivative:
f'(x) = 5x^{51} + 0
f'(x) = 5x^{4}
now the 2^{nd} derivative:
f''(x) = 4(5x^{41})
f''(x) = 20x^{3}
now the 3^{rd} derivative:
f'''(x) = 3(20x^{31})
f'''(x) = 60x^{2}
now the 4^{th} derivative:
f^{(4)}(x) = 2(60x^{21})
f^{(4)}(x) = 120x
finally the 5^{th} derivative:
f^{(5)}(x) = 120 x^{11}
f^{(5)}(x) = 120 x^{0} = 120

C^{2} is a constant, so it vanishes in the first
derivative.




Example:

Let's continue the previous example and find the the
6^{th} derivative.
f^{(6)}(x) = 0.

The derivative of the constant 120 is 0.
In general, if the order of the derivative is larger than
the degree of the polynomial, then that derivative vanishes,
that is, it becomes zero.




Example:

find y^{(4)} (θ) = cos(θ)
y^{'} (θ) = sin(θ)
y'' (θ) = cos(θ)
y''' (θ) = (sin(θ)) = sin(θ)
y^{(4)} (θ) =
cos(θ)

This is an interesting example. Every 4 derivatives it
repeats.
For example:
y^{(7)} (θ) = sin(θ)
y^{(22)} (θ) = cos(θ)




Example:

Evaluate :

As always, with grouping symbols, work from the inside
out.
First find the derivative of (x^{2} + 1).
Now simplify.
Finally find the derivative of 2x^{2}.
Simplify once more.




Example:

find
first derivative:

Here we use negative exponents to avoid fractions.
Finally we simplify by rewriting without negative
exponents.




(advanced)



Example:

Find:
Now look at the multipliers. The first derivative has the
multiplier n. The second has n(n1). The last factor (n1) is
(n  2 + 1).
The third has n(n1)(n2) and the last factor is (n  3 +
1).
So we would expect the n^{th} derivative to have
the last factor (n – n + 1) = 1, which taken with the other factors
in the product becomes n!.

Find derivatives one after the other and look for a
pattern.
The 1 vanishes in the first derivative (x^{n1}
– 0).
The first derivative has the exponent( n – 1).
The 2^{nd} derivative has the exponent (n –
2).
The 3^{rd} derivative has the exponent (n –
3).
We would expect the 4^{th} derivative to have the
exponent (n – 4).
Finally the n^{th} derivative has the
exponent (n  n) = 0.

(advanced)



Example:

Find:

As before find the derivatives one at a time and look for
a pattern.
We need to use (and you should first review) the
Reciprocal Rule for derivatives here.
Look for patterns as we go from derivative to
derivative.
Note:
At the start of each derivative I will move the numerator
in front of the fraction.





The sequence is:
n=1, n=2, n=3, n=4
The numerators are factorials 1!, 2!, 3!, 4!,
...
We need alternating signs so (1)^{n}
will work as a factor for each term.
The powers of (x1) are an arithmetic sequence
with distance 1. To match our exponents with n we use the
exponent (n+1). So the n^{th} term is:
which is
Important note:
Recall with polynomials a derivative whose
order is larger than the degree of the polynomial causes
the derivative to vanish.
Here we have negative exponent 1. As we take
derivatives, the result has an exponent in the denominator
whose magnitude is one more than the order of the derivative. We can take
derivatives all day long in this case without the derivative
vanishing, ever.

Examples involving higher order derivatives can be found in the topic
Derivative
Applications.