HigherOrder Derivatives


HigherOrder Derivatives of an Explicit Function
Let the function y = f(x) have a finite derivative
f '(x) in a certain interval (a, b),
i.e. the derivative f '(x) is also a function in this interval.
If this function is differentiable, we can find the second derivative
of the original function f, which is denoted as
Similarly, if f '' exists and is differentiable, we can calculate the
third derivative of the function f:
The derivatives of higher order (if they exist) are defined as
Thus, the notion of the nth order derivative is introduced inductively by sequential calculation of n derivatives
starting from the first order derivative. Transition to the next higherorder derivative is performed using
the recurrence formula
In some cases, we can derive a general formula for the derivative of an arbitrary nth order without computing intermediate derivatives.
Some examples are considered below.
Note that the following linear relationships can be used for finding higherorder derivatives:
HigherOrder Derivatives of an Implicit Function
The nth order derivative of an implicit function can be found by sequential ( n times)
differentiation of the equation F(x, y) = 0. At each step, after appropriate substitutions
and transformations, we can obtain an explicit expression for the derivative, which depends only on the variables x and y, i.e.
the derivatives have the form
HigherOrder Derivatives of a Parametric Function
Consider a function y = f(x) given parametrically
by the equations
The first derivative of this function is given by
Differentiating once more with respect to x, we find the second derivative:
Similarly, we define the derivatives of the third and higher order:

Example 1


Find y'', if y(x) = x ln x.
Solution.
Calculate the first derivative using the
product rule:
Now we can find the second derivative:

Example 2


Find the second derivative of the function .
Solution.
We start with first derivative:
Differentiate again to find the second derivative:

Example 3


Calculate y'' for the parabola equation y^{2} = 4x.
Solution.
By implicit differentiation,
Differentiating again and using the product rule, we obtain
Multiply both sides by y^{ 2} :
Since yy' = 2 and, hence,
(yy' )^{2} = 4, we can write the last equation as
Then

Example 4


Given the function y = (2x + 1)^{3}(x − 1).
Find all derivatives of the nth order from n = 1 to
n = 5.
Solution.
First we convert the given function into a polynomial:
Now we successively calculate the derivatives from 1st to 5th order:

Example 5


Find the nth order derivative of the natural logarithm function y = ln x.
Solution.
We calculate several successive derivatives of the given function:
We see that the derivative of an arbitrary nth order is given by
A rigorous justification of this formula can be obtained using the method of mathematical induction.

Example 6


Find all derivatives of the sine function.
Solution.
Calculate a few derivatives starting from the first one:
Obviously, the norder derivative is expressed by the formula
A rigorous proof of this formula can be done by induction.

Example 7


Find all derivatives of the cosine function.
Solution.
Analogously to Example 6, we find the first few derivatives of the cosine function:
Clearly that the next 5th order derivative coincides with the first derivaive, the 6th with the 2nd and so on.
Thus, the nth order derivative of the cosine function is described by the formula

Example 8


Find all derivatives of the function .
Solution.
We find a few first derivatives:
This is enough to detect the general pattern:

Example 9


Determine the second derivative of the function .
Solution.
Differentiating as a composite function,
we find the first derivative:
Here we can represent  x as x sign x where
Then
We now calculate the second derivative differentiating the previous expression as a
quotient of two functions:

Example 10


Find the third derivative of the function .
Solution.
Differentiate successively the given function:

Example 11


Find the second derivative of the implicitly defined function x^{2} + y^{2} = R^{2}
( canonical equation of a circle).
Solution.
Given that y is a function of x and differentiating both sides of the equation, we find the first derivative:
We differentiate this expression again:
Substituting the first derivative y' into this formula, we have

Example 12


Find the nth order derivative of the function y = 3^{2x+1}.
Solution.
We calculate successively several derivatives, starting from the first one.
It follows from here that the nth order derivative is given by
A rigorous proof can be carried out by induction. Clearly that this formula is valid for n = 1.
Suppose that it is true for n = k :
Then for n = k + 1 we have
i.e. this formula holds for n = k + 1.
Consequently, it is true for any natural number n.

Example 13


Find the nth order derivative of the power function y = x^{m}
where m is a real number.
Solution.
We calculate several first derivatives:
Hence it is easy to establish a general expression for the nth order derivative:
We prove this by induction. Obviously, this formula is valid for n = 1.
Assuming that it is true for the degree n, we differentiate it again and find the derivative of
(n + 1)th order:
As can be seen, the derivative of (n + 1)th order is expressed by the same formula
as the nth order derivative (only the number n is replaced by n + 1).
Consequently, the resulting expression is valid for any positive integer value of n ( n is the order of the derivative).
Note that the exponent m, generally speaking, is a real number. If we consider only the natural values of m,
then the formula for the derivative can be written in a more compact form:
where n ≤ m. All other derivatives of order n > m
are equal to zero.

Example 14


Find the nth order derivative of the square root
y = √x.
Solution.
We use the result of Example 13, where the nth order derivative of the power function with an arbitrary real exponent m
is derived. In this case we have
Then the derivative is written as
For n = 1, the derivative is
Provided n ≥ 2, the product of odd numbers in square brackets
can be written in terms of the
double factorial:
Hence for n ≥ 2, the nth order derivative is expressed by the general formula
In particular,

Example 15


Find the nth order derivative of the cube root .
Solution.
The first derivative of the cube root is given by
Next, we use the general formula for the nth order derivative of the power function
y = x^{m} (Example 13):
In our case m = 1/3. Consequently, the derivative is as follows:
where n ≥ 2. Specifically, the second and third derivatives of the cube root
are expressed by the formulas

Example 16


Find the first and second derivatives with respect to x of the function given parametrically:
Solution.
Compute the first derivative y'_{x} by the
formula:
Consequently,
Differentiating again, we find the second derivative with respect to x:
Calculate separately the derivative in the numerator:
Since the derivative in the denominator is equal to
we obtain the following expression for the second derivative of the original function:

Example 17


Given the equation of an ellipse
in parametric form:
where a, b are semiaxes of the ellipse, t is a parameter. Find the first, second and third derivatives
of the function y with respect to x.
Solution.
Differentiating the given parametric function successively, we obtain:

Example 18


Find the third derivative of the function given by the equation
x^{2} + 3xy + y^{2} = 1.
Solution.
One time differentiation with respect to x leads to the following expression:
We differentiate the last expression again considering y as a composite function:
Substitute the explicit expression for the first derivative y':
Since x^{2} + 3xy + y^{2} = 1,
we obtain the following expression for the second derivative y'':
Similarly, differentiating once more, we get the third derivative:
We substitute again the first derivative and find:

Example 19


Find the second derivative of the function given by the equation
x + y = e^{ x − y}.
Solution.
Differentiating both sides by x, we obtain:
Continuing the differentiation, we find the second derivative:
Substitute the expression for the first derivative:
We now use the original equation, according to which
As a result, we obtain the following expression for the derivative y'':

Example 20


A curve is described by the implicit equation
xy = 2x^{3} − y^{2}
and passes through the point (1, 1). Find the first and second derivatives
at this point.
Solution.
We calculate the first derivative:
Differentiating again, we find the value of the second derivative at the given point:
Substitute the known values of x, y, y':

