Calculus

Differentiation of Functions

Differentiation of Functions Logo

Higher-Order Differentials

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.

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:

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

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:

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

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

Click or tap a problem to see the solution.

Example 1

Find the differential \({d^4}y\) of the function \(y = {x^4}.\)

Example 2

Find the differential \({d^4}y\) of the function \(y = {x^5}.\)

Example 3

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

Example 4

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

Example 5

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

Example 6

Find \({d^3}y\) of the function \(y = x\ln \large{\frac{1}{x}}\normalsize.\)

Example 7

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

Example 8

Find the differential \({d^3}y\) of the function \(y = \large{\frac{1}{{\sqrt x }}}\normalsize.\)

Example 9

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

Example 10

Find the \(n\)th-order differential of the function \(y = {2^x}.\)

Example 11

Find the differential \({d^3}y\) of the function \(y = {x^2}{e^x}.\)

Example 12

Find the differential \({d^5}y\) of the function \(y = \ln {x^2}.\)

Example 13

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

Example 14

Find the differential \({d^3}y\) of the function \(y = x\sinh x.\)

Example 15

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

Example 16

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

Example 17

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 18

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 19

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 20

Find the second order differential of the function \(y = {u^3},\) where \(u = u\left( x \right).\)

Example 21

Find the second order differential of the function \(y = {e^{2u}},\) where \(u = u\left( x \right).\)

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:

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

Hence,

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

Example 3.

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

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

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:

\[
{y’ = {\left( {{x^2}\cos 2x} \right)^\prime } }
= {{\left( {{x^2}} \right)^\prime }\cos 2x }+{ {x^2}{\left( {\cos 2x} \right)^\prime } }
= {2x\cos 2x }+{ {x^2} \cdot \left( { – 2\sin 2x} \right) }={ 2x\cos 2x }-{ 2{x^2}\sin 2x,}
\]

\[
{y^{\prime\prime} }={ {\left( {2x\cos 2x – 2{x^2}\sin 2x} \right)^\prime } }
= {2{\left( {x\cos 2x – {x^2}\sin 2x} \right)^\prime } }
= {2\left[ {x’\cos 2x + x{{\left( {\cos 2x} \right)}^\prime } }\right.}\kern0pt{-\left.{ {{\left( {{x^2}} \right)}^\prime }\sin 2x – {x^2}{{\left( {\sin 2x} \right)}^\prime }} \right] }
= {2\left[ {\cos 2x – 2x\sin 2x }\right.}-{\left.{ 2x\sin 2x – 2{x^2}\cos 2x} \right] }
= {\left( {2 – 2{x^2}} \right)\cos 2x }-{ 4x\sin 2x.}
\]

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 \(y = x\ln \large{\frac{1}{x}}\normalsize.\)

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

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

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Problems 1-6
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Problems 7-21