# Calculus

Differentiation of Functions# Higher-Order Differentials

Problems 1-2

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

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:

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

In the general case, the differential of an arbitrary order \(n\) is given by

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

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

### 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 \(x\):

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

Consequently,

or in short form:

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

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

### ✓ 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

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,

### 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

Problems 1-2

Problems 3-11