# Calculus

## Differentiation of Functions # Higher-Order Differentials

## Concept of Higher-Order Differentials

We consider a function y = f (x), which is differentiable in the interval (a, b). The first-order differential of the function at the point x (a, b) is defined by the formula

$dy = f'\left( x \right)dx.$

It can be seen that the differential dy depends on two quantities - the variable x (through the derivative y' = f '(x)) and the differential of the independent variable dx.

Let us fix the increment $$dx$$, i.e. we assume that $$dx$$ is constant. Then the differential $$dy$$ becomes a function only of the variable $$x,$$ for which we can also define the differential by taking the same differential $$dx$$ as the increment $$\Delta x.$$ As a result, we obtain the second differential or differential of the second order, which is denoted as $${d^2}y$$ or $${d^2}f\left( x \right).$$ Thus, by definition:

${d^2}y = d\left( {dy} \right) = d\left[ {f'\left( x \right)dx} \right] = df'\left( x \right)dx = f^{\prime\prime}\left( x \right)dxdx = f^{\prime\prime}\left( x \right){\left( {dx} \right)^2}.$

It is commonly denoted $${\left( {dx} \right)^2} = d{x^2}.$$ Therefore, we get:

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

In the same way, we can establish that the third differential or differential of the third order has the form

${d^3}y = f^{\prime\prime\prime}\left( x \right)d{x^3}.$

In the general case, the differential of an arbitrary order $$n$$ is given by

${d^n}y = {f^{\left( n \right)}}\left( x \right)d{x^n},$

which can be rigorously proved by mathematical induction. This formula leads in particular to the following expression for the $$n$$th order derivative:

$f^{\left( n \right)}\left( x \right) = \frac{{{d^n}y}}{{d{x^n}}}.$

Note that for the linear function $$y = ax + b,$$ the second and subsequent higher-order differentials are zero. Indeed,

${d^2}\left( {ax + b} \right) = \left( {ax + b} \right)^{\prime\prime} d{x^2} = {0 \cdot d{x^2} = 0,} \ldots , {d^n}\left( {ax + b} \right) = 0.$

In this case, it is obvious that

${d^n}x = 0\;\;\text{for}\;\;n \gt 1.$

## Properties of Higher-Order Differentials

Let the functions $$u$$ and $$v$$ have the $$n$$th order derivatives. Then the following properties are valid:

• $${{d^n}\left( {\alpha u + \beta v} \right) }$$ $$={ \alpha {d^n}u + \beta {d^n}v;}$$
• $${{d^n}\left( {uv} \right) }$$ $$={ \sum\limits_{i = 0}^n {C_n^i{d^{n - i}}u{d^i}v}.}$$

The last equality follows directly from the Leibniz formula.

## Higher Order Differential of a Composite Function

Consider now the composition of two functions such that $$y = f\left( u \right)$$ and $$u = g\left( x \right).$$ In this case, $$y$$ is a composite function of the independent variable $$x:$$

$y = f\left( {g\left( x \right)} \right).$

The first differential of $$y$$ can be written as

$dy = {\left[ {f\left( {g\left( x \right)} \right)} \right]^\prime }dx = f'\left( {g\left( x \right)} \right)g'\left( x \right)dx.$

Compute the second differential $${d^2}y$$ (assuming $$dx$$ is constant by definition). Using the product rule, we obtain:

${d^2}y = {\left[ {f'\left( {g\left( x \right)} \right)g'\left( x \right)} \right]^\prime }d{x^2} = \left[ {f^{\prime\prime}\left( {g\left( x \right)} \right){{\left( {g'\left( x \right)} \right)}^2} + f'\left( {g\left( x \right)} \right)g^{\prime\prime}\left( x \right)} \right]d{x^2} = f^{\prime\prime}\left( {g\left( x \right)} \right){\left( {g'\left( x \right)dx} \right)^2} + f'\left( {g\left( x \right)} \right)g^{\prime\prime}\left( x \right)d{x^2}.$

Take into account that

$g'\left( x \right)dx = du\;\;\text{and}\;\;\;g^{\prime\prime}\left( x \right)d{x^2} = {d^2}u.$

Consequently,

${d^2}y = f^{\prime\prime}\left( u \right)d{u^2} + f'\left( u \right){d^2}u$

or in short form:

${d^2}y = y^{\prime\prime}d{u^2} + y'{d^2}u.$

In the same way, we can obtain the expression for the third order differential of a composite function:

${d^3}y = f^{\prime\prime\prime}\left( u \right)d{u^3} + 3f^{\prime\prime}\left( u \right)du{d^2}u + f^\prime\left( u \right){d^3}u.$

It follows from the above that the higher order differentials $${d^2}y,{d^3}y, \ldots ,{d^n}y$$ are generally not invariant.

## Solved Problems

### Example 1.

Find the differential $${d^4}y$$ of the function $y = {x^4}.$

Solution.

Determine the $$4$$th derivative:

$y^\prime = \left( {{x^4}} \right)^\prime = 4{x^3};$
$y^{\prime\prime} = \left( {4{x^3}} \right)^\prime = 4 \cdot 3{x^2} = 12{x^2};$
$y^{\prime\prime\prime} = \left( {12{x^2}} \right)^\prime = 12 \cdot 2x = 24x;$
${y^{\left( 4 \right)}} = \left( {24x} \right)^\prime = 24.$

Then the differential $${d^4}y$$ is given by

${d^4}y = 24d{x^4}.$

### Example 2.

Find the differential $${d^4}y$$ of the function $y = {x^5}.$

Solution.

The $$4$$th order differential is given by

${d^4}y = f^{\left( 4 \right)}\left( x \right)d{x^4} = \left( {{x^5}} \right)^{\left( 4 \right)} d{x^4}.$

We find the fourth derivative of this function by successive differentiation:

$\left( {{x^5}} \right)' = 5{x^4},\;\;\;\left( {{x^5}} \right)^{\prime\prime} = \left( {5{x^4}} \right)' = 20{x^3},\;\;\;\left( {{x^5}} \right)^{\prime\prime\prime} = \left( {20{x^3}} \right)' = 60{x^2},\;\;\;\left( {{x^5}} \right)^{\left( 4 \right)} = \left( {60{x^2}} \right)' = 120x.$

Hence,

${d^4}y = 120x\,d{x^4}.$

### Example 3.

Given the function $y = \sqrt{x}.$ Find $${d^2}y.$$

Solution.

We calculate the first and second derivatives of the given function:

$y' = f'\left( x \right) = \left( {\sqrt{x}} \right)^\prime = \left( {{x^{\frac{1}{3}}}} \right)^\prime = \frac{1}{3}{x^{\frac{1}{3} - 1} = \frac{1}{3}{x^{ - \frac{2}{3}}},}\;\;\;y^{\prime\prime} = f^{\prime\prime}\left( x \right) = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right){x^{ - \frac{5}{3}}} = - \frac{2}{{9\sqrt{{{x^5}}}}}.$

Then the second-order differential is written as

${d^2}y = f^{\prime\prime}\left( x \right)d{x^2} = - \frac{{2d{x^2}}}{{9\sqrt{{{x^5}}}}}.$

### Example 4.

Find the differential $${d^5}y$$ of the function $y = \sin 2x.$

Solution.

It is known that the $$n$$th-order derivative of the sine function has the form

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

One can show that the $$n$$th-order derivative of the function $$y = \sin 2x$$ is given by

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

Hence, the $$5$$th-order derivative is written as

$y^{\left( 5 \right)} = \left( {\sin 2x} \right)^{\left( 5 \right)} = {2^5}\sin \left( {2x + \frac{{5\pi }}{2}} \right) = 32\sin \left( {2x + 2\pi + \frac{\pi }{2}} \right) = 32\sin \left( {2x + \frac{\pi }{2}} \right) = 32\cos 2x.$

This yields

${d^5}y = 32\cos 2x\,d{x^5}.$

### Example 5.

Find the second differential of the function $y = {x^2}\cos 2x.$

Solution.

Determine the second derivative of this function:

$y' = \left( {{x^2}\cos 2x} \right)^\prime = {\left( {{x^2}} \right)^\prime }\cos 2x + {x^2}{\left( {\cos 2x} \right)^\prime } = 2x\cos 2x + {x^2} \cdot \left( { - 2\sin 2x} \right) = 2x\cos 2x - 2{x^2}\sin 2x,$
$y^{\prime\prime} = \left( {2x\cos 2x - 2{x^2}\sin 2x} \right)^\prime = 2{\left( {x\cos 2x - {x^2}\sin 2x} \right)^\prime } = 2\left[ {x'\cos 2x + x{{\left( {\cos 2x} \right)}^\prime {\left( {{x^2}} \right)}^\prime }\sin 2x - {x^2}{{\left( {\sin 2x} \right)}^\prime }} \right] = 2\left[ {\cos 2x - 2x\sin 2x - 2x\sin 2x - 2{x^2}\cos 2x} \right] = \left( {2 - 2{x^2}} \right)\cos 2x - 4x\sin 2x.$

Then the second-order differential is written in the form:

${d^2}y = y^{\prime\prime}d{x^2} = \left[ {\left( {2 - 2{x^2}} \right)\cos 2x - 4x\sin 2x} \right]d{x^2}.$

### Example 6.

Find $${d^3}y$$ of the function $y = x\ln \frac{1}{x}.$

Solution.

The third order differential is given by

${d^3}y = y^{\prime\prime\prime}\left( x \right) d{x^3}.$

We differentiate the given function successively:

$y^\prime = \left( {x\ln \frac{1}{x}} \right)^\prime = x \cdot \left( {\frac{1}{{\frac{1}{x}}}} \right) \cdot \left( {\frac{1}{x}} \right)^\prime + 1 \cdot \ln \frac{1}{x} = {x^2} \cdot \left( { - \frac{1}{{{x^2}}}} \right) + \ln \frac{1}{x} = \ln \frac{1}{x} - 1;$
$y^{\prime\prime} = \left( {\ln \frac{1}{x} - 1} \right)^\prime = \left( {\frac{1}{{\frac{1}{x}}}} \right) \cdot \left( {\frac{1}{x}} \right)^\prime = x \cdot \left( { - \frac{1}{{{x^2}}}} \right) = - \frac{1}{x};$
$y^{\prime\prime\prime} = \left( { - \frac{1}{x}} \right)^\prime = \frac{1}{{{x^2}}}.$

Hence

${d^3}y = y^{\prime\prime\prime}\left( x \right)d{x^3} = \frac{{d{x^3}}}{{{x^2}}}.$

### Example 7.

Find the $$4$$th order differential of the function $y = \sqrt {2x + 1}.$

Solution.

Differentiating this function successively, we find its $$4$$th derivative:

$\require{cancel} y' = \left( {\sqrt {2x + 1} } \right)^\prime = \frac{1}{{2\sqrt {2x + 1} }} \cdot {\left( {2x + 1} \right)^\prime } = \frac{\cancel{2}}{{\cancel{2}\sqrt {2x + 1} }} = \frac{1}{{\sqrt {2x + 1} }} = \left( {2x + 1} \right)^{ - \frac{1}{2}},$
$y^{\prime\prime} = \left[ {{{\left( {2x + 1} \right)}^{ - \frac{1}{2}}}} \right]^\prime = - \frac{1}{2}{\left( {2x + 1} \right)^{ - \frac{3}{2}}} \cdot {\left( {2x + 1} \right)^\prime } = - \left( {2x + 1} \right)^{ - \frac{3}{2}};$
$y^{\prime\prime\prime} = \left[ { - {{\left( {2x + 1} \right)}^{ - \frac{3}{2}}}} \right]^\prime = \frac{3}{2}{\left( {2x + 1} \right)^{ - \frac{5}{2}}} \cdot {\left( {2x + 1} \right)^\prime } = 3{\left( {2x + 1} \right)^{ - \frac{5}{2}}};$
$y^{\left( 4 \right)} = \left[ {3{{\left( {2x + 1} \right)}^{ - \frac{5}{2}}}} \right]^\prime = 3 \cdot \left( { - \frac{5}{2}} \right){\left( {2x + 1} \right)^{ - \frac{7}{2}}} \cdot {\left( {2x + 1} \right)^\prime } = - \frac{{15}}{2}{\left( {2x + 1} \right)^{ - \frac{7}{2}}} \cdot 2 = - \frac{{15}}{{\sqrt {{{\left( {2x + 1} \right)}^7}} }}.$

Consequently, the $$4$$th order differential is given by

${d^4}y = {y^{\left( 4 \right)}}d{x^4} = - \frac{{15\,d{x^4}}}{{\sqrt {{{\left( {2x + 1} \right)}^7}} }}.$

### Example 8.

Find the differential $${d^3}y$$ of the function $y = \frac{1}{{\sqrt x }}.$

Solution.

Calculate the third derivative:

$y^\prime = \left( {\frac{1}{{\sqrt x }}} \right)^\prime = \left( {{x^{ - \frac{1}{2}}}} \right)^\prime = - \frac{1}{2}{x^{ - \frac{3}{2}}};$
$y^{\prime\prime} = \left( { - \frac{1}{2}{x^{ - \frac{3}{2}}}} \right)^\prime = - \frac{1}{2} \cdot \left( { - \frac{3}{2}} \right){x^{ - \frac{5}{2}}} = \frac{3}{4}{x^{ - \frac{5}{2}}};$
$y^{\prime\prime\prime} = \left( {\frac{3}{4}{x^{ - \frac{5}{2}}}} \right)^\prime = \frac{3}{4} \cdot \left( { - \frac{5}{2}} \right){x^{ - \frac{7}{2}}} = - \frac{{15}}{8}{x^{ - \frac{7}{2}}} = - \frac{{15}}{{8\sqrt {{x^7}} }}.$

Hence, the third order differential is given by

${d^3}y = y^{\prime\prime\prime}\left( x \right)d{x^3} = - \frac{{15d{x^3}}}{{8\sqrt {{x^7}} }}.$

See more problems on Page 2.