Higher-Order Derivatives

Higher-Order 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 y = f (x), which is denoted by various notations as

$f^{\prime\prime} = \left({f^\prime}\right)^\prime = {\left( {\frac{{dy}}{{dx}}} \right)^\prime } = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{{d^2}y}}{{d{x^2}}}.$

Similarly, if f '' exists and is differentiable, we can calculate the third derivative of the function f (x):

$f^{\prime\prime\prime} = \frac{{{d^3}y}}{{d{x^3}}} = y^{\prime\prime\prime}.$

The result of taking the derivative $$n$$ times is called the $$n$$th derivative of $$f\left( x \right)$$ with respect to $$x$$ and is denoted as

$\frac{{{d^n}f}}{{d{x^n}}} = \frac{{{d^n}y}}{{d{x^n}}}\;\;\;\left( {\text{in Leibnitz's notation}} \right),$
${f^{\left( n \right)}}\left( x \right) = {y^{\left( n \right)}}\left( x \right)\;\;\;\left( {\text{in Lagrange's notation}} \right).$

Thus, the notion of the $$n$$th order derivative is introduced inductively by sequential calculation of $$n$$ derivatives starting from the first order derivative. Transition to the next higher-order derivative is performed using the recurrence formula

$y^{\left( n \right)} = \left( {{y^{\left( {n - 1} \right)}}} \right)^\prime.$

In some cases, we can derive a general formula for the derivative of an arbitrary $$n$$th order without computing intermediate derivatives. Some examples are considered below.

Note that the following linear relationships can be used for finding higher-order derivatives:

$\left( {u + v} \right)^{\left( n \right)} = {u^{\left( n \right)}} + {v^{\left( n \right)}},\;\;\;{\left( {Cu} \right)^{\left( n \right)}} = C{u^{\left( n \right)}},\;\;C = \text{const}.$

Higher-Order Derivatives of an Implicit Function

The $$n$$th order derivative of an implicit function can be found by sequential ($$n$$ times) differentiation of the equation $$F\left( {x,y} \right) = 0.$$ 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

$y' = {f_1}\left( {x,y} \right),\;\;\;y^{\prime\prime} = {f_2}\left( {x,y} \right), \ldots,\;\;\;{y^{\left( n \right)}} = {f_n}\left( {x,y} \right).$

Higher-Order Derivatives of a Parametric Function

Consider a function $$y = f\left( x \right)$$ given parametrically by the equations

\left\{ \begin{aligned} x &= x\left( t \right) \\ y &= y\left( t \right) \end{aligned} \right..

The first derivative of this function is given by

$y' = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}}.$

Differentiating once more with respect to $$x,$$ we find the second derivative:

$y^{\prime\prime} = {y^{\prime\prime}_{xx}} = \frac{{{\left( {{y'_x}} \right)}'_t}}{{{x'_t}}}.$

Similarly, we define the derivatives of the third and higher order:

$y^{\prime\prime\prime} = {y^{\prime\prime\prime}_{xxx}} = \frac{{{{\left( {{y^{\prime\prime}_{xx}}} \right)}'_t}}}{{{x'_t}}}, \ldots,\;\;y^{\left( n \right)} = y_{\underbrace {xx \ldots x}_n}^{\left( n \right)} = \frac{{{{\left( {y_{\underbrace {xx \ldots x}_{n - 1}}^{\left( {n - 1} \right)}} \right)}'_t}}}{{{x'_t}}}.$

Solved Problems

Click or tap a problem to see the solution.

Example 1

Given the function $y = {\left( {2x + 1} \right)^3}\left( {x - 1} \right).$ Find all derivatives of the $$n$$th order from $$n = 1$$ to $$n = 5.$$

Example 2

Find the $$n$$th order derivative of the natural logarithm function $y = \ln x.$

Example 3

Find the $$n$$th derivative of the function $y = \ln \left( {ax + b} \right).$

Example 4

Find the $$n$$th derivative of the function $y = x\ln x$ at $$x = 1.$$

Example 5

Find all derivatives of the sine function.

Example 6

Find the $$n$$th derivative of the function $y = \sin ax.$

Example 7

Find all derivatives of the cosine function.

Example 8

Find the $$n$$th order derivative of the function $y = {\sin ^2}x.$

Example 1.

Given the function $y = {\left( {2x + 1} \right)^3}\left( {x - 1} \right).$ Find all derivatives of the $$n$$th order from $$n = 1$$ to $$n = 5.$$

Solution.

First we convert the given function into a polynomial:

$y = {\left( {2x + 1} \right)^3}\left( {x - 1} \right) = \left( {8{x^3} + 12{x^2} + 6x + 1} \right)\cdot \left( {x - 1} \right) = \color{blue}{8{x^4}} + \color{red}{12{x^3}} + \color{maroon}{6{x^2}} + \color{green}x - \color{red}{8{x^3}} - \color{maroon}{12{x^2}} - \color{green}{6x} - \color{coral}1 = \color{blue}{8{x^4}} + \color{red}{4{x^3}} - \color{maroon}{6{x^2}} - \color{green}{5x} - \color{coral}1.$

Now we successively calculate the derivatives from $$1$$st to $$5$$th order:

$y' = \left( {8{x^4} + 4{x^3} - 6{x^2} - 5x - 1} \right)^\prime = 32{x^3} + 12{x^2} - 12x - 5,$
$y^{\prime\prime} = {\left( {y'} \right)^\prime } = {\left( {32{x^3} + 12{x^2} - 12x - 5} \right)^\prime } = 96{x^2} + 24x - 12,$
$y^{\prime\prime\prime} = {\left( {y^{\prime\prime}} \right)^\prime } = {\left( {96{x^2} + 24x - 12} \right)^\prime } = 192x + 24,$
${y^{\left( 4 \right)}} = {\left( {y^{\prime\prime\prime}} \right)^\prime } = {\left( {192x + 24} \right)^\prime } = 192,$
${y^{\left( 5 \right)}} = {\left( {{y^{\left( 4 \right)}}} \right)^\prime } = {\left( {192} \right)^\prime } = 0.$

Example 2.

Find the $$n$$th order derivative of the natural logarithm function $y = \ln x.$

Solution.

We calculate several successive derivatives of the given function:

$y' = \left( {\ln x} \right)^\prime = \frac{1}{x},$
$y^{\prime\prime} = {\left( {y'} \right)^\prime } = {\left( {\frac{1}{x}} \right)^\prime } = {\left( {{x^{ - 1}}} \right)^\prime } = - {x^{ - 2}} = - \frac{1}{{{x^2}}},$
$y^{\prime\prime\prime} = {\left( {y^{\prime\prime}} \right)^\prime } = {\left( { - \frac{1}{{{x^2}}}} \right)^\prime } = 2{x^{ - 3}} = \frac{2}{{{x^3}}},$
${y^{\left( 4 \right)}} = {\left( {y^{\prime\prime\prime}} \right)^\prime } = {\left( {\frac{2}{{{x^3}}}} \right)^\prime } = - 6{x^{ - 4}} = - \frac{6}{{{x^4}}},$
${y^{\left( 5 \right)}} = {\left( {{y^{\left( 4 \right)}}} \right)^\prime } = {\left( { - \frac{6}{{{x^4}}}} \right)^\prime } = 24{x^{ - 5}} = \frac{{24}}{{{x^5}}}.$

We see that the derivative of an arbitrary $$n$$th order is given by

$y^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!}}{{{x^n}}}.$

A rigorous justification of this formula can be obtained using the method of mathematical induction.

Example 3.

Find the $$n$$th derivative of the function $y = \ln \left( {ax + b} \right).$

Solution.

Let's calculate the few first derivatives using the chain and power rules:

$y^\prime = \left( {\ln \left( {ax + b} \right)} \right)^\prime = \frac{1}{{ax + b}} \cdot \left( {ax + b} \right)^\prime = \frac{a}{{ax + b}};$
$y^{\prime\prime} = \left( {\frac{a}{{ax + b}}} \right)^\prime = \left( {a{{\left( {ax + b} \right)}^{ - 1}}} \right)^\prime = - 1 \cdot {a^2}{\left( {ax + b} \right)^{ - 2}} = - \frac{{1 \cdot {a^2}}}{{{{\left( {ax + b} \right)}^2}}};$
$y^{\prime\prime\prime} = \left( { - 1 \cdot {a^2}{{\left( {ax + b} \right)}^{ - 2}}} \right)^\prime = - 1 \cdot \left( { - 2} \right) \cdot {a^3}{\left( {ax + b} \right)^{ - 3}} = 2!\,{a^3}{\left( {ax + b} \right)^{ - 3}} = \frac{{2!\,{a^3}}}{{{{\left( {ax + b} \right)}^3}}};$
${y^{\left( 4 \right)}} = \left( {2!{a^3}{{\left( {ax + b} \right)}^{ - 3}}} \right)^\prime = 2! \cdot \left( { - 3} \right) \cdot {a^4}{\left( {ax + b} \right)^{ - 4}} = - 3!\,{a^4}{\left( {ax + b} \right)^{ - 4}} = - \frac{{3!\,{a^4}}}{{{{\left( {ax + b} \right)}^4}}}.$

We can easily detect the common pattern, so the $$n$$th derivative is given by

$y^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!\,{a^n}}}{{{{\left( {ax + b} \right)}^n}}}.$

Example 4.

Find the $$n$$th derivative of the function $y = x\ln x$ at $$x = 1.$$

Solution.

The first two derivatives are written as

$y^\prime = \left( {x\ln x} \right)^\prime = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1;$
$y^{\prime\prime} = \left( {\ln x + 1} \right)^\prime = \frac{1}{x} = {x^{ - 1}}.$

We can continue differentiating using the power rule:

$y^{\prime\prime\prime} = \left( {{x^{ - 1}}} \right)^\prime = - 1 \cdot {x^{ - 2}} = - \frac{1}{{{x^2}}};$
${y^{\left( 4 \right)}} = \left( { - 1 \cdot {x^{ - 2}}} \right)^\prime = 2!\,{x^{ - 3}} = \frac{{2!}}{{{x^3}}};$
${y^{\left( 5 \right)}} = \left( {2!\,{x^{ - 3}}} \right)^\prime = 2! \cdot \left( { - 3} \right) \cdot {x^{ - 4}} = - 3!\,{x^{ - 4}} = - \frac{{3!}}{{{x^4}}}.$

We see that the $$n$$th derivative $$\left(\text{at } {n \ge 2} \right)$$ has the form

$y^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^n}\left( {n - 1} \right)!}}{{{x^{n - 1}}}}.$

Substituting $$x = 1$$ yields:

${y^{\left( n \right)}}\left( 1 \right) = \frac{{{{\left( { - 1} \right)}^n}\left( {n - 1} \right)!}}{{{1^{n - 1}}}} = {\left( { - 1} \right)^n}\left( {n - 1} \right)!$

Example 5.

Find all derivatives of the sine function.

Solution.

We calculate a few derivatives starting from the first one:

$y' = {\left( {\sin x} \right)^\prime } = \cos x = \sin \left( {x + \frac{\pi }{2}} \right),$
$y^{\prime\prime} = {\left( {\cos x} \right)^\prime } = - \sin x = \sin \left( {x + 2 \cdot \frac{\pi }{2}} \right),$
$y^{\prime\prime\prime} = {\left( { - \sin x} \right)^\prime } = - \cos x = \sin \left( {x + 3 \cdot \frac{\pi }{2}} \right),$
${y^{IV}} = {\left( { - \cos x} \right)^\prime } = \sin x = \sin \left( {x + 4 \cdot \frac{\pi }{2}} \right).$

Obviously, the $$n$$-order derivative is expressed by the formula

${y^{\left( n \right)}} = {\left( {\sin x} \right)^{\left( n \right)}} = \sin \left( {x + \frac{{n\pi }}{2}} \right).$

A rigorous proof of this formula can be done by induction.

Example 6.

Find the $$n$$th derivative of the function $y = \sin ax.$

Solution.

Differentiating successively and using the trig cofunction identities, we have

$y^\prime = \left( {\sin ax} \right)^\prime = a\cos ax = a\sin \left( {ax + \frac{\pi }{2}} \right);$
$y^{\prime\prime} = \left( {a\cos ax} \right)^\prime = - {a^2}\sin ax = {a^2}\sin \left( {ax + 2 \cdot \frac{\pi }{2}} \right);$
$y^{\prime\prime\prime} = \left( { - {a^2}\sin ax} \right)^\prime = - {a^3}\cos ax = {a^3}\sin \left( {ax + 3 \cdot \frac{\pi }{2}} \right);$
${y^{\left( 4 \right)}} = \left( { - {a^3}\cos ax} \right)^\prime = {a^4}\sin ax = {a^4}\sin \left( {ax + 2\pi } \right) = {a^4}\sin \left( {ax + 4 \cdot \frac{\pi }{2}} \right).$

Then, the $$n$$th derivative is written in the form

$y^{\left( n \right)} = {a^n}\sin \left( {ax + \frac{{\pi n}}{2}} \right).$

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:

$y' = {\left( {\cos x} \right)^\prime } = - \sin x = \cos \left( {x + \frac{\pi }{2}} \right),$
$y^{\prime\prime} = {\left( { - \sin x} \right)^\prime } = - \cos x = \cos \left( {x + 2 \cdot \frac{\pi }{2}} \right),$
$y^{\prime\prime\prime} = {\left( { - \cos x} \right)^\prime } = \sin x = \cos \left( {x + 3 \cdot \frac{\pi }{2}} \right),$
${y^{IV}} = {\left( {\sin x} \right)^\prime } = \cos x = \cos \left( {x + 4 \cdot \frac{\pi }{2}} \right).$

Clearly that the next $$5$$th order derivative coincides with the first derivaive, the $$6$$th with the $$2$$nd and so on. Thus, the $$n$$th order derivative of the cosine function is described by the formula

${y^{\left( n \right)}} = {\left( {\cos x} \right)^{\left( n \right)}} = \cos \left( {x + \frac{{n\pi }}{2}} \right).$

Example 8.

Find the $$n$$th order derivative of the function $y = {\sin ^2}x.$

Solution.

Let's calculate the first few derivatives:

$y^\prime = \left( {{{\sin }^2}x} \right)^\prime = 2\sin x\cos x = \sin 2x;$
$y^{\prime\prime} = \left( {\sin 2x} \right)^\prime = 2\cos 2x = 2\sin \left( {2x + \frac{\pi }{2}} \right);$
$y^{\prime\prime\prime} = \left( {2\cos 2x} \right)^\prime = - 4\sin 2x = {2^2}\sin \left( {2x + \frac{\pi }{2} \cdot 2} \right);$
${y^{\left( 4 \right)}} = \left( { - 4\sin 2x} \right)^\prime = - 8\cos 2x = {2^3}\sin \left( {2x + \frac{\pi }{2} \cdot 3} \right).$

We used here cofunction identities:

$\sin \left( {\alpha + \frac{\pi }{2}} \right) = \cos \alpha ;$
$\sin \left( {\alpha + \pi } \right) = - \sin \alpha ;$
$\sin \left( {\alpha + \frac{{3\pi }}{2}} \right) = - \cos \alpha .$

It is easy to detect the following pattern:

${y^{\left( n \right)}} = {2^{n - 1}}\sin \left( {2x + \frac{{\pi n}}{2}} \right),$

where $$n = 1,2,3, \ldots$$

See more problems on Page 2.