Higher-Order Differentials
Solved Problems
Example 9.
Find \({d^3}y\) of the function \[y = {x^2}\ln x.\]
Solution.
The third-order differential is defined by the relationship:
\[{d^3}y = f^{\prime\prime\prime}\left( x \right)d{x^3}.\]
We find the third-order derivative \(f^{\prime\prime\prime}\left( x \right)\) by the Leibniz formula:
\[f^{\prime\prime\prime}\left( x \right) = \left( {{x^2}\ln x} \right)^{\prime\prime\prime} = \sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}} 3\\ i \end{array}} \right){{\left( {\ln x} \right)}^{\left( {3 - i} \right)}}{{\left( {{x^2}} \right)}^{\left( i \right)}}} .\]
The derivatives in this series are given by
\[\left( {\ln x} \right)' = \frac{1}{x},\;\;\;\left( {\ln x} \right)^{\prime\prime} = \left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{{x^2}}},\;\;\;\left( {\ln x} \right)^{\prime\prime\prime} = \left( { - \frac{1}{{{x^2}}}} \right)^\prime = \frac{2}{{{x^3}}};\]
\[\left( {{x^2}} \right)' = 2x,\;\;\;\left( {{x^2}} \right)^{\prime\prime} = \left( {2x} \right)^\prime = 2,\;\;\;{\left( {{x^2}} \right)^{\prime\prime\prime}} = 0.\]
As a result, we obtain
\[
f^{\prime\prime\prime}\left( x \right) = \left( {{x^2}\ln x} \right)^{\prime\prime\prime}
= \left( {\begin{array}{*{20}{c}}
3\\
0
\end{array}} \right) \cdot \frac{2}{{{x^3}}} \cdot {x^2} + \left( {\begin{array}{*{20}{c}}
3\\
1
\end{array}} \right) \cdot \left( { - \frac{1}{{{x^2}}}} \right) \cdot 2x + \left( {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right) \cdot \frac{1}{x} \cdot 2 + \left( {\begin{array}{*{20}{c}}
3\\
3
\end{array}} \right) \cdot \ln x \cdot 0 = \frac{2}{x} - \cancel{\frac{6}{x}} + \cancel{\frac{6}{x}} = \frac{2}{x}.
\]
Then the differential is
\[{d^3}y = f^{\prime\prime\prime}\left( x \right)d{x^3} = \frac{{2\,d{x^3}}}{x}.\]
Example 10.
Find the \(n\)th-order differential of the function \[y = {2^x}.\]
Solution.
We calculate the first few derivatives:
\[y^\prime = \left( {{2^x}} \right)^\prime = {2^x}\ln 2;\]
\[y^{\prime\prime} = \left( {{2^x}\ln 2} \right)^\prime = {2^x}{\left( {\ln 2} \right)^2};\]
\[y^{\prime\prime\prime} = \left( {{2^x}{{\left( {\ln 2} \right)}^2}} \right)^\prime = {2^x}{\left( {\ln 2} \right)^3}.\]
So the \(n\)th-order derivative is given by
\[y^{\left( n \right)} = {2^x}{\left( {\ln 2} \right)^n}.\]
Then, the \(n\)th-order differential is written in the form
\[{d^n}y = {2^x}{\left( {\ln 2} \right)^n}d{x^n}.\]
Example 11.
Find the differential \({d^3}y\) of the function \[y = {x^2}{e^x}.\]
Solution.
Differentiate the given function successively:
\[y^\prime = \left( {{x^2}{e^x}} \right)^\prime = \left( {{x^2}} \right)^\prime{e^x} + {x^2}\left( {{e^x}} \right)^\prime = 2x{e^x} + {x^2}{e^x} = \left( {{x^2} + 2x} \right){e^x};\]
\[y^{\prime\prime} = \left( {\left( {{x^2} + 2x} \right){e^x}} \right)^\prime = \left( {{x^2} + 2x} \right)^\prime{e^x} + \left( {{x^2} + 2x} \right)\left( {{e^x}} \right)^\prime = \left( {2x + 2} \right){e^x} + \left( {{x^2} + 2x} \right){e^x} = \left( {{x^2} + 4x + 2} \right){e^x};\]
\[y^{\prime\prime\prime} = \left( {\left( {{x^2} + 4x + 2} \right){e^x}} \right)^\prime = \left( {{x^2} + 4x + 2} \right)^\prime{e^x} + \left( {{x^2} + 4x + 2} \right)\left( {{e^x}} \right)^\prime = \left( {2x + 4} \right){e^x} + \left( {{x^2} + 4x + 2} \right){e^x} = \left( {{x^2} + 6x + 6} \right){e^x}.\]
Hence, the third order differential is given by
\[{d^3}y = \left( {{x^2} + 6x + 6} \right){e^x}d{x^3}.\]
Example 12.
Find the differential \({d^5}y\) of the function \[y = \ln {x^2}.\]
Solution.
We compute the fifth derivative by differentiating the function successively:
\[y^\prime = \left( {\ln {x^2}} \right)^\prime = \frac{1}{{{x^2}}} \cdot \left( {{x^2}} \right)^\prime = \frac{{2x}}{{{x^2}}} = \frac{2}{x};\]
\[y^{\prime\prime} = \left( {\frac{2}{x}} \right)^\prime = 2\left( {{x^{ - 1}}} \right)^\prime = - 2{x^{ - 2}} = - \frac{2}{{{x^2}}};\]
\[y^{\prime\prime\prime} = \left( { - 2{x^{ - 2}}} \right)^\prime = 4{x^{ - 3}} = \frac{4}{{{x^3}}};\]
\[{y^{\left( 4 \right)}} = \left( {4{x^{ - 3}}} \right)^\prime = - 12{x^{ - 4}} = - \frac{{12}}{{{x^4}}};\]
\[{y^{\left( 5 \right)}} = \left( { - 12{x^{ - 4}}} \right)^\prime = 48{x^{ - 5}} = \frac{{48}}{{{x^5}}}.\]
Hence
\[{d^5}y = \frac{{48d{x^5}}}{{{x^5}}}.\]
Example 13.
Find the \(5\)th order differential of the function \[y = x\cos x.\]
Solution.
First we calculate the fifth-order derivative. For this we use the Leibniz formula:
\[y^{\left( 5 \right)} = \left( {x\cos x} \right)^{\left( 5 \right)} = \sum\limits_{i = 0}^5 {\left( {\begin{array}{*{20}{c}} 5\\ i \end{array}} \right){{\left( {\cos x} \right)}^{\left( {5 - i} \right)}}{x^{\left( i \right)}}} .\]
The derivative of the cosine function of an arbitrary kth order is determined by the following formula (see the Higher-Order Derivatives page):
\[\left( {\cos x} \right)^{\left( k \right)} = \cos \left( {x + \frac{{\pi k}}{2}} \right).\]
Since the \(k\)th order derivative of \(x\) is equal to zero for \(k \gt 1,\) the stated series is limited to two terms and has the following form:
\[
{y^{\left( 5 \right)}} = \sum\limits_{i = 0}^5 {\left( {\begin{array}{*{20}{c}}
5\\
i
\end{array}} \right){{\left( {\cos x} \right)}^{\left( {5 - i} \right)}}{x^{\left( i \right)}}} = \left( {\begin{array}{*{20}{c}}
5\\
0
\end{array}} \right){\left( {\cos x} \right)^{\left( 5 \right)}}x + \left( {\begin{array}{*{20}{c}}
5\\
1
\end{array}} \right){\left( {\cos x} \right)^{\left( 4 \right)}} = x\cos \left( {x + \frac{{5\pi }}{2}} \right) + 5\cos \left( {x + \frac{{4\pi }}{2}} \right) = - x\sin x + 5\cos x = 5\cos x - x\sin x.\]
Thus, the \(5\)th order differential of the given function is written as
\[{d^5}y = {y^{\left( 5 \right)}}d{x^5} = \left( {5\cos x - x\sin x} \right)d{x^5}.\]
Example 14.
Find the differential \({d^3}y\) of the function \[y = x\sinh x.\]
Solution.
Calculate the third derivative of the function:
\[y^\prime = \left( {x\sinh x} \right)^\prime = x^\prime\sinh x + x\left( {\sinh x} \right)^\prime = \sinh x + x\cosh x;\]
\[y^{\prime\prime} = \left( {\sinh x + x\cosh x} \right)^\prime = \left( {\sinh x} \right)^\prime + x^\prime\cosh x + x\left( {\cosh x} \right)^\prime = \cosh x + \cosh x + x\sinh x = 2\cosh x + x\sinh x;\]
\[y^{\prime\prime\prime} = \left( {2\cosh x + x\sinh x} \right)^\prime = \left( {2\cosh x} \right)^\prime + x^\prime\sinh x + x\left( {\sinh x} \right)^\prime = 2\sinh x + \sinh x + x\cosh x = 3\sinh x + x\cosh x.\]
Hence, the \(3\)rd-order differential is given by
\[{d^3}y = \left( {3\sinh x + x\cosh x} \right)d{x^3}.\]
Example 15.
Find the differential \({d^8}y\) of the function \[y = {e^x}\sin x.\]
Solution.
The indicated differential is written as
\[{d^8}y = {f^{\left( 8 \right)}}\left( x \right)d{x^8}.\]
Since the function \(y\) is the product of the functions \({e^x}\) and \(\sin x,\) then the derivative of the \(8\)th order can be found by Leibniz formula:
\[
y^{\left( 8 \right) = f^{\left( 8 \right)}}\left( x \right)
=\left( {{e^x}\sin x} \right)^{\left( 8 \right)}
= \sum\limits_{i = 0}^8 {\left( {\begin{array}{*{20}{c}}
8\\
i
\end{array}} \right){{\left( {\sin x} \right)}^{\left( {8 - i} \right)}}{{\left( {{e^x}} \right)}^{\left( i \right)}}} .\]
The derivatives contained in the series are as follows (see the Higher-Order Derivatives page):
\[{\left( {\sin x} \right)^{\left( k \right)}} = \sin \left( {x + \frac{{\pi k}}{2}} \right),\;\; \Rightarrow \left( {\sin x} \right)^{\left( {8 - i} \right)} = \sin \left( {x + \frac{{\pi \left( {8 - i} \right)}}{2}} \right) = \sin \left( {x + 4\pi - \frac{{\pi i}}{2}} \right) = \sin \left( {x - \frac{{\pi i}}{2}} \right);\]
\[\left( {{e^x}} \right)^{\left( i \right)}= {e^x}.\]
Then the derivative \({y^{\left( 8 \right)}}\) is given by
\[y^{\left( 8 \right)} = \sum\limits_{i = 0}^8 {\left( {\begin{array}{*{20}{c}} 8\\ i \end{array}} \right){{\left( {\sin x} \right)}^{\left( {8 - i} \right)}}{{\left( {{e^x}} \right)}^{\left( i \right)}}} = {e^x}\left[ {\left( {\begin{array}{*{20}{c}} 8\\ 0 \end{array}} \right)\sin x} \right. + \left( {\begin{array}{*{20}{c}} 8\\ 1 \end{array}} \right)\sin \left( {x - \frac{\pi }{2}} \right) + \left( {\begin{array}{*{20}{c}} 8\\ 2 \end{array}} \right)\sin \left( {x - \pi } \right) + \left( {\begin{array}{*{20}{c}} 8\\ 3 \end{array}} \right)\sin \left( {x - \frac{{3\pi }}{2}} \right) + \left( {\begin{array}{*{20}{c}} 8\\ 4 \end{array}} \right)\sin \left( {x - 2\pi } \right) + \left( {\begin{array}{*{20}{c}} 8\\ 5 \end{array}} \right)\sin \left( {x - \frac{{5\pi }}{2}} \right) + \left( {\begin{array}{*{20}{c}} 8\\ 6 \end{array}} \right)\sin \left( {x - 3\pi } \right) + \left( {\begin{array}{*{20}{c}} 8\\ 7 \end{array}} \right)\sin \left( {x - \frac{{7\pi }}{2}} \right) + \left. \left( {\begin{array}{*{20}{c}} 8\\ 8 \end{array}} \right)\sin \left( {x - 4\pi } \right) \right] = {{e^x}\left[ {\sin x} \right. - \frac{{8!}}{{7!\,1!}}\cos x - \frac{{8!}}{{6!\,2!}}\sin x } + {\frac{{8!}}{{5!\,3!}}\cos x + \frac{{8!}}{{4!\,4!}}\sin x } - {\frac{{8!}}{{3!\,5!}}\cos x - \frac{{8!}}{{2!\,6!}}\sin x } + {\frac{{8!}}{{1!\,7!}}\cos x + \left. \sin x \right] } = {{e^x} \left[ {\color{blue}{\sin x}} \right. - \cancel{\color{red}{8\cos x}} - \color{blue}{28\sin x} + \cancel{\color{red}{56\cos x}} } + {\color{blue}{70\sin x} - \cancel{\color{red}{56\cos x}} - \color{blue}{28\sin x} } + {\cancel{\color{red}{8\cos x}} + \left. \color{blue}{\sin x} \right]} = 16{e^x}\sin x. \]
Hence, the \(8\)th order differential is written in the form
\[{d^8}y = f^{\left( 8 \right)}\left( x \right)d{x^8} = \left( {{e^x}\sin x} \right)^{\left( 8 \right)} d{x^8} = 16{e^x}\sin x\,d{x^8}.\]
Example 16.
Given the function \[y = \sin x\cosh x.\] Find \({d^3}y.\)
Solution.
We calculate the third derivative of the function using the Leibniz rule:
\[
y^{\prime\prime\prime} = \left( {\sin x\cosh x} \right)^{\prime\prime\prime} = \sum\limits_{i = 0}^3 {\left( {\begin{array}{*{20}{c}}
3\\
i
\end{array}} \right){{\left( {\sin x} \right)}^{\left( {3 - i} \right)}}{{\left( {\cosh x} \right)}^{\left( i \right)}}} .\]
Write the derivatives contained in this series:
\[\left( {\sin x} \right)' = \cos x,\;\;\;\left( {\sin x} \right)^{\prime\prime} = \left( {\cos x} \right)' = - \sin x,\;\;\;\left( {\sin x} \right)^{\prime\prime\prime} = \left( { - \sin x} \right)' = - \cos x;\]
\[\left( {\cosh x} \right)' = \sinh x,\;\;\;\left( {\cosh x} \right)^{\prime\prime} = \left( {\sinh x} \right)' = \cosh x,\;\left( {\cosh x} \right)^{\prime\prime\prime} = \left( {\cosh x} \right)' = \sinh x.\]
Hence, the derivative \(y^{\prime\prime\prime}\) is expressed by the formula
\[
y^{\prime\prime\prime} = \left( {\sin x\cosh x} \right)^{\prime\prime\prime}
= \left( {\begin{array}{*{20}{c}}
3\\
0
\end{array}} \right){\left( {\sin x} \right)^{\prime\prime\prime}}\cosh x + \left( {\begin{array}{*{20}{c}}
3\\
1
\end{array}} \right){\left( {\sin x} \right)^{\prime\prime}} {\left( {\cosh x} \right)'} + \left( {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right){\left( {\sin x} \right)'}{\left( {\cosh x} \right)^{\prime\prime}} + \left( {\begin{array}{*{20}{c}}
3\\
3
\end{array}} \right)\sin x{\left( {\cosh x} \right)'} = - \color{blue}{\cos x\cosh x } - \color{red}{3\sin x \sinh x } + \color{blue}{3\cos x\cosh x } + \color{red}{\sin x \sinh x } = \color{blue}{2\cos x\cosh x} - \color{red}{2\sin x \sinh x}.\]
Accordingly, the differential \({d^3}y\) is written as
\[{d^3}y = y^{\prime\prime\prime}d{x^3} = \left( {2\cos x\cosh x - 2\sin x\sinh x} \right)d{x^3}.\]
Example 17.
Find the second differential \({d^2}y\) of the astroid defined by the equation
\[{x^{\frac{2}{3}}} + {y^{\frac{2}{3}}} = {R^{\frac{2}{3}}}.\]
Solution.
We compute successively the first and second derivatives of the function \(y\left( x \right)\) describing the astroid. Differentiating both sides of the implicit equation with respect to \(x\), we have:
\[\left( {{x^{\frac{2}{3}} + y^{\frac{2}{3}}}} \right)^\prime = \left( {{R^{\frac{2}{3}}}} \right)^\prime ,\;\; \Rightarrow \frac{2}{3}{x^{ - \frac{1}{3}}} + \frac{2}{3}{y^{ - \frac{1}{3}}}y' = 0,\;\; \Rightarrow {x^{ - \frac{1}{3}}} + {y^{ - \frac{1}{3}}}y' = 0,\;\; \Rightarrow {y^{ - \frac{1}{3}}}y' = - x^{ - \frac{1}{3}},\;\; \Rightarrow y' = - {\left( {\frac{x}{y}} \right)^{ - \frac{1}{3}}} = - \left( {\frac{y}{x}} \right)^{\frac{1}{3}}.\]
Differentiate again, given that \(y\) is the function of \(x:\)
\[y^{\prime\prime} = \left[ { - {{\left( {\frac{y}{x}} \right)}^{\frac{1}{3}}}} \right]^\prime = - \frac{1}{3}{\left( {\frac{y}{x}} \right)^{ - \frac{2}{3}}} \cdot {\left( {\frac{y}{x}} \right)^\prime } = - \frac{1}{3}{\left( {\frac{x}{y}} \right)^{\frac{2}{3}}} \cdot \frac{{y'x - yx'}}{{{x^2}}} = - \frac{{{x^{\frac{2}{3}}}}}{{3{y^{\frac{2}{3}}}}} \cdot \frac{{y'x - y}}{{{x^2}}}.\]
Substitute the expression for the first derivative \(y'\) found above:
\[y^{\prime\prime} = - \frac{{{x^{\frac{2}{3}}}}}{{3{y^{\frac{2}{3}}}}} \cdot \frac{{\left( { - {{\left( {\frac{y}{x}} \right)}^{\frac{1}{3}}}} \right)x - y}}{{{x^2}}} = - \frac{{{x^{\frac{2}{3}}}}}{{3{y^{\frac{2}{3}}}}} \cdot \frac{{\left( { - {y^{\frac{1}{3}}}{x^{\frac{2}{3}}} - y} \right)}}{{{x^2}}} = \frac{{{x^{\frac{2}{3}}}{y^{\frac{1}{3}}}\left( {{x^{\frac{2}{3}} + y^{\frac{2}{3}}}} \right)}}{{3{y^{\frac{2}{3}}}{x^2}}} = \frac{{{R^{\frac{2}{3}}}}}{{3{y^{\frac{1}{3}}}{x^{\frac{4}{3}}}}}.\]
Then the second differential is given by
\[{d^2}y = y^{\prime\prime}d{x^2} = \frac{{{R^{\frac{2}{3}}}d{x^2}}}{{3{y^{\frac{1}{3}}}{x^{\frac{4}{3}}}}}.\]
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 \({d^2}y.\)
Solution.
We determine the second-order differential by the formula
\[{d^2}y = y^{\prime\prime}\left( x \right)d{x^2}.\]
Find the second derivative \(y''\left( x \right).\) The first derivative is given by
\[y'\left( x \right) = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {{t^3} - 2t} \right)}^\prime }}}{{{{\left( {{t^2} + t - 1} \right)}^\prime }}} = \frac{{3{t^2} - 2}}{{2t + 1}}.\]
Then the second derivative can be expressed as follows:
\[y^{\prime\prime}\left( x \right) = y^{\prime\prime}_{xx} = \left( {{y'_x}} \right)'_x = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( {\frac{{3{t^2} - 2}}{{2t + 1}}} \right)'}}}}{{{{\left( {{t^2} + t - 1} \right)'}}}} = \frac{{\frac{{{{\left( {3{t^2} - 2} \right)'}}\left( {2t + 1} \right) - \left( {3{t^2} - 2} \right){{\left( {2t + 1} \right)'}}}}{{{{\left( {2t + 1} \right)}^2}}}}}{{2t + 1}} = \frac{{6t \cdot \left( {2t + 1} \right) - \left( {3{t^2} - 2} \right) \cdot 2}}{{{{\left( {2t + 1} \right)}^3}}} = \frac{{12{t^2} + 6t - 6{t^2} + 4}}{{{{\left( {2t + 1} \right)}^3}}} = \frac{{6{t^2} + 6t + 4}}{{{{\left( {2t + 1} \right)}^3}}}.\]
Calculate the differential \(d{x^2}:\)
\[d{x^2} = \left( {dx} \right)^2 = \left( {d\left( {{t^2} + t - 1} \right)} \right)^2 = \left( {\left( {2t + 1} \right)dt} \right)^2 = \left( {2t + 1} \right)^2 \left( {dt} \right)^2 = {\left( {2t + 1} \right)^2}d{t^2}.\]
Thus, the differential of the \(2\)nd order of the original function is given by
\[{d^2}y = y^{\prime\prime}\left( x \right)d{x^2} = \frac{{6{t^2} + 6t + 4}}{{{{\left( {2t + 1} \right)}^3}}} \cdot \left( {2t + 1} \right)^2 d{t^2} = \frac{{6{t^2} + 6t + 4}}{{2t + 1}}d{t^2}.\]
Example 19.
The equation of the upper semicircle centered at the origin with radius \(R\) in parametric form is given by
\[x = R\cos t,\;y = R\sin t,\] where \(0 \lt t \lt \pi.\) Find the third differential \({d^3}y\) of this function.
We calculate successively the derivatives of the \(1\)st, \(2\)nd and \(3\)rd order with respect to the variable \(x:\)
\[y' = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {R\sin t} \right)}^\prime }}}{{{{\left( {R\cos t} \right)}^\prime }}} = \frac{{\cancel{R}\cos t}}{{ - \cancel{R}\sin t}} = - \cot t,\]
\[y^{\prime\prime} = y^{\prime\prime}_{xx} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( { - \cot t} \right)'}}}}{{{{\left( {R\cos t} \right)'}}}} = \frac{{ - \left( { - \frac{1}{{{{\sin }^2}t}}} \right)}}{{ - R\sin t}} = - \frac{1}{{R\,{{\sin }^3}t}} = - \frac{1}{R}{\csc ^3}t,\]
\[y^{\prime\prime\prime} = y^{\prime\prime\prime}_{xxx} = \frac{{{{\left( {{y^{\prime\prime}_{xx}}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( { - \frac{1}{R}{{\csc }^3}t} \right)'}}}}{{{{\left( {R\cos t} \right)'}}}} = \frac{{\left( { - \frac{1}{R}} \right) \cdot 3{{\csc }^2}t \cdot {{\left( {\csc t} \right)'}}}}{{\left( { - R\sin t} \right)}} = \frac{{3{{\csc }^2}t \cdot \left( { - \cot t} \right) \cdot \csc t}}{{{R^2}\sin t}} = - \frac{3}{{{R^2}}}{\csc ^4}t\cot t.\]
Express the differential \(d{x^3}\) in terms of the variable \(t:\)
\[d{x^3} = \left( {dx} \right)^3 = \left( {d\left( {R\cos t} \right)} \right)^3 = \left( { - R\sin t\,dt} \right)^3 = - {R^3}{\sin ^3}t{\left( {dt} \right)^3} = - {R^3}{\sin ^3}t\,d{t^3}.\]
It follows from here that
\[{d^3}y = y^{\prime\prime\prime}\left( x \right)d{x^3} = y^{\prime\prime\prime}_{xxx} d{x^3} = \left( { - \frac{3}{{{R^2}}}} \right){\csc ^4}t\cot t \cdot \left( { - {R^3}{{\sin }^3}t} \right)d{t^3} = 3R\csc t\cot t \cdot \frac{1}{{{{\sin }^3}t}} \cdot {\sin ^3}t\,d{t^3} = 3R\csc t\cot t\,d{t^3}.\]
Example 20.
Find the second order differential of the function \[y = {u^3},\] where \(u = u\left( x \right).\)
Solution.
The first differential of the composite function is written as
\[dy = d\left( {{u^3}} \right) = \left( {{u^3}} \right)^\prime du = 3{u^2}du.\]
Compute the second order differential:
\[{d^2}y = d\left( {dy} \right) = d\left( {3{u^2}du} \right) = d\left( {3{u^2}} \right) \cdot du + 3{u^2} \cdot d\left( {du} \right) = 6udu \cdot du + 3{u^2} \cdot {d^2}u = 6u{\left( {du} \right)^2} + 3{u^2}{d^2}u = 6ud{u^2} + 3{u^2}{d^2}u.\]
Example 21.
Find the second order differential of the function \[y = {e^{2u}},\] where \(u = u\left( x \right).\)
Solution.
The first differential is written in the form
\[dy = d\left( {{e^{2u}}} \right) = \left( {{e^{2u}}} \right)^\prime du = 2{e^{2u}}du.\]
Determine the second order differential keeping in mind that \(y\) is a composite function:
\[{d^2}y = d\left( {dy} \right) = d\left( {2{e^{2u}}du} \right) = d\left( {2{e^{2u}}} \right)du + 2{e^{2u}}d\left( {du} \right) = 2\left( {{e^{2u}}} \right)^\prime dudu + 2{e^{2u}}{d^2}u = 4{e^{2u}}{\left( {du} \right)^2} + 2{e^{2u}}{d^2}u = 4{e^{2u}}d{u^2} + 2{e^{2u}}{d^2}u = 2{e^{2u}}\left( {2d{u^2} + {d^2}u} \right).\]
