Select Page

# Calculus

Differentiation of Functions

# Higher-Order Differentials

Page 1
Problems 1-2
Page 2
Problems 3-11

### Concept of Higher-Order Differentials

We consider a function $$y = f\left( x \right)$$, which is differentiable in the interval $$\left( {a,b} \right).$$ The first-order differential of the function at the point $$x \in \left( {a,b} \right)$$ 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 $$y = f’\left( x \right)$$) 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:

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

Its 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

${{d^n}x = 0\;\;}\kern-0.3pt{\text{for}\;\;n \gt 1.}$

### 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.

### Second 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:

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

Take into account that

${g’\left( x \right)dx = du\;\;\text{and}\;\;\;}\kern-0.3pt{g^{\prime\prime}\left( x \right)d{x^2} = {d^2}u.}$

Consequently,

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

or in short form:

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

The obtained formulas show that unlike the first differential, the second differential does not have the property of form invariance. The additional term $$y'{d^2}u$$ appears in the formula for the second differential. However, in the case of a linear function $${u = g\left( x \right) }$$ $$={ ax + b}$$ its second differential is equal to

${{d^2}u = {d^2}\left( {ax + b} \right) }={ 0.}$

Therefore, in this case the invariance of the form of the second differential $${d^2}u$$ holds.

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

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

### ✓Example 2

Given the function $$y = \sqrt[3]{x}$$. Find $${d^2}y.$$

### ✓Example 3

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

### ✓Example 4

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

### ✓Example 5

Find $${d^3}y$$ of the function $$y = {x^2}\ln x.$$

### ✓Example 6

Find the $$5$$th order differential of the function $$y = x\cos x.$$

### ✓Example 7

Find the differential $${d^8}y$$ of the function $$y = {e^x}\sin x.$$

### ✓Example 8

Given the function $$y = \sin x\cosh x.$$ Find $${d^3}y.$$

### ✓Example 9

Find the second differential $${d^2}y$$ of the astroid defined by the equation

${x^{\large\frac{2}{3}\normalsize}} + {y^{\large\frac{2}{3}\normalsize}} = {R^{\large\frac{2}{3}\normalsize}}.$

### ✓Example 10

The function is given in parametric form by the equations

\left\{ \begin{aligned} x &= {t^2} + t – 1 \\ y &= {t^3} – 2t \end{aligned} \right..

Find the second-order differential $${d^2}y.$$

### ✓Example 11

The equation of the upper semicircle centered at the origin with radius $$R$$ in parametric form is given by

${x = R\cos t,\;\;}\kern-0.3pt{y = R\sin t,\;\;\;}\kern-0.3pt {\text{where}\;\;0 \lt t \lt \pi .}$

Find the third differential $${d^3}y$$ of this function.

### Example 1.

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:

${{\left( {{x^5}} \right)’} = 5{x^4},\;\;\;}\kern-0.3pt {{\left( {{x^5}} \right)^{\prime\prime}} = {\left( {5{x^4}} \right)’} = 20{x^3},\;\;\;}\kern-0.3pt {{\left( {{x^5}} \right)^{\prime\prime\prime}} = {\left( {20{x^3}} \right)’} = 60{x^2},\;\;\;}\kern-0.3pt {{\left( {{x^5}} \right)^{\left( 4 \right)}} = {\left( {60{x^2}} \right)’} = 120x.}$

Hence,

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

### Example 2.

Given the function $$y = \sqrt[3]{x}$$. Find $${d^2}y.$$

#### Solution.

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

${y’ = f’\left( x \right) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } = {\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{\large\frac{1}{3}\normalsize – 1}} = {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}},}\;\;\;}\kern-0.3pt {y^{\prime\prime} = {f^{\prime\prime}\left( x \right)} = {\frac{1}{3} \cdot \left( { – \frac{2}{3}} \right){x^{ – \large\frac{5}{3}\normalsize}} }= { – \frac{2}{{9\sqrt[\large 3\normalsize]{{{x^5}}}}}.}}$

Then the second-order differential is written as

${{d^2}y = f^{\prime\prime}\left( x \right)d{x^2} }={ – \frac{{2d{x^2}}}{{9\sqrt[\large 3\normalsize]{{{x^5}}}}}.}$
Page 1
Problems 1-2
Page 2
Problems 3-11