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