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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


fifth derivative


nth derivative


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 f4(x) and f4(x) is f(x) raised to the 4th 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:


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.


find f''(x) for f(x) = ½ x2 + x + 1

the first derivative:

f'(x) = 2( ½ x2-1 )+ 1x1-1 + 0

f'(x) = x1 + x0

f'(x) = x + 1

now the second derivative:

f''(x) = 1(x1-1) + 0

f''(x) = x0 = 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.


find f(5)(x) for f(x) = x5 + C2

the first derivative:

f'(x) = 5x5-1 + 0

f'(x) = 5x4

now the 2nd derivative:

f''(x) = 4(5x4-1)

f''(x) = 20x3

now the 3rd derivative:

f'''(x) = 3(20x3-1)

f'''(x) = 60x2

now the 4th derivative:

f(4)(x) = 2(60x2-1)

f(4)(x) = 120x

finally the 5th derivative:

f(5)(x) = 120 x1-1

f(5)(x) = 120 x0 = 120

C2 is a constant, so it vanishes in the first derivative.


Let's continue the previous example and find the the 6th 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.


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)


Evaluate :

As always, with grouping symbols, work from the inside out.

First find the derivative of (x2 + 1).

Now simplify.

Finally find the derivative of 2x2.

Simplify once more.



first derivative:

Here we use negative exponents to avoid fractions.

Finally we simplify by rewriting without negative exponents.




Now look at the multipliers. The first derivative has the multiplier n. The second has n(n-1). The last factor (n-1) is (n - 2 + 1).

The third has n(n-1)(n-2) and the last factor is (n - 3 + 1).

So we would expect the nth 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 (xn-1 – 0).

The first derivative has the exponent( n – 1).

The 2nd derivative has the exponent (n – 2).

The 3rd derivative has the exponent (n – 3).

We would expect the 4th derivative to have the exponent (n – 4).

Finally the nth derivative has the exponent
(n - n) = 0.




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.


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 (x-1) are an arithmetic sequence with distance 1. To match our exponents with n we use the exponent (n+1). So the nth 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.


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