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 \[y = \sqrt[3]{x}.\] Find \({d^2}y.\)
Solution.
We calculate the first and second derivatives of the given function:
\[y' = f'\left( x \right) = \left( {\sqrt[3]{x}} \right)^\prime = \left( {{x^{\frac{1}{3}}}} \right)^\prime = \frac{1}{3}{x^{\frac{1}{3} - 1} = \frac{1}{3}{x^{ - \frac{2}{3}}},}\;\;\;y^{\prime\prime} = f^{\prime\prime}\left( x \right) = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right){x^{ - \frac{5}{3}}} = - \frac{2}{{9\sqrt[3]{{{x^5}}}}}.\]
Then the second-order differential is written as
\[{d^2}y = f^{\prime\prime}\left( x \right)d{x^2} = - \frac{{2d{x^2}}}{{9\sqrt[3]{{{x^5}}}}}.\]
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.
