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   Higher-Order Derivatives
Higher-Order Derivatives of an Explicit Function
Let the function \(y = f\left( x \right)\) have a finite derivative \(f'\left( x \right)\) in a certain interval \(\left( {a,b} \right),\) i.e. the derivative \(f'\left( x \right)\) is also a function in this interval. If this function is differentiable, we can find the second derivative of the original function \(f\left( x \right)\), which is denoted as \[f'' = f' = {\left( {\frac{{dy}}{{dx}}} \right)^\prime } = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{{d^2}y}}{{d{x^2}}}.\] Similarly, if \(f''\) exists and is differentiable, we can calculate the third derivative of the function \(f\left( x \right)\): \[f''' = \frac{{{d^3}y}}{{d{x^3}}} = y'''.\] The derivatives of higher order (if they exist) are defined as \[ {{f^{\left( 4 \right)}} = \frac{{{d^4}y}}{{d{x^4}}} = {y^{\left( 4 \right)}} = {\left( {{f^{\left( 3 \right)}}} \right)^\prime },} \ldots , {{f^{\left( n \right)}} = \frac{{{d^n}y}}{{d{x^n}}} = {y^{\left( n \right)}} = {\left( {{f^{\left( {n - 1} \right)}}} \right)^\prime }.} \] Thus, the notion of the \(n\)th order derivative is introduced inductively by sequential calculation of \(n\) derivatives starting from the first order derivative. Transition to the next higher-order derivative is performed using the recurrence formula \[{y^{\left( n \right)}} = {\left( {{y^{\left( {n - 1} \right)}}} \right)^\prime }.\] In some cases, we can derive a general formula for the derivative of an arbitrary \(n\)th order without computing intermediate derivatives. Some examples are considered below.

Note that the following linear relationships can be used for finding higher-order derivatives: \[ {{\left( {u + v} \right)^{\left( n \right)}} = {u^{\left( n \right)}} + {v^{\left( n \right)}},}\;\;\; {{\left( {Cu} \right)^{\left( n \right)}} = C{u^{\left( n \right)}},\;C = \text{const}.} \]
Higher-Order Derivatives of an Implicit Function
The \(n\)th order derivative of an implicit function can be found by sequential (\(n\) times) differentiation of the equation \(F\left( {x,y} \right) = 0.\). At each step, after appropriate substitutions and transformations, we can obtain an explicit expression for the derivative, which depends only on the variables \(x\) and \(y\), i.e. the derivatives have the form \[ {y' = {f_1}\left( {x,y} \right),}\;\; {y'' = {f_2}\left( {x,y} \right), \ldots,}\;\; {{y^{\left( n \right)}} = {f_n}\left( {x,y} \right).} \]
Higher-Order Derivatives of a Parametric Function
Consider a function \(y = f\left( x \right)\) given parametrically by the equations \[ \left\{ \begin{aligned} x &= x\left( t \right) \\ y &= y\left( t \right) \end{aligned} \right.. \] The first derivative of this function is given by \[y' = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}}.\] Differentiating once more with respect to \(x\), we find the second derivative: \[y'' = {y''_{xx}} = \frac{{{\left( {{y'_x}} \right)}'_t}}{{{x'_t}}}.\] Similarly, we define the derivatives of the third and higher order: \[ {y''' = {y'''_{xxx}} = \frac{{{{\left( {{y''_{xx}}} \right)}'_t}}}{{{x'_t}}}, \ldots,}\; {{y^{\left( n \right)}} = y_{\underbrace {xx \ldots x}_n}^{\left( n \right)} = \frac{{{{\left( {y_{\underbrace {xx \ldots x}_{n - 1}}^{\left( {n - 1} \right)}} \right)}'_t}}}{{{x'_t}}}.} \]
   Example 1
Find \(y''\), if \(y = x\ln x.\)

Solution.
Calculate the first derivative using the product rule: \[ {y' = \left( {x\ln x} \right)' } ={ x' \cdot \ln x + x \cdot {\left( {\ln x} \right)^\prime } } ={ 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1.} \] Now we can find the second derivative: \[ {y'' = {\left( {\ln x + 1} \right)^\prime } } = {\frac{1}{x} + 0 = \frac{1}{x}.} \]
   Example 2
Find the second derivative of the function \(y = \sqrt[\large 4\normalsize]{{x + 1}}.\)

Solution.
We start with first derivative: \[ y' = {\left( {\sqrt[\large 4\normalsize]{{x + 1}}} \right)^\prime } = {\left[ {{{\left( {x + 1} \right)}^{\large\frac{1}{4}\normalsize}}} \right]^\prime } = \frac{1}{4}{\left( {x + 1} \right)^{ - \large\frac{3}{4}\normalsize}} = \frac{1}{{4\sqrt[\large 4\normalsize]{{{{\left( {x + 1} \right)}^3}}}}}. \] Differentiate again to find the second derivative: \[ {y'' = {\left( {y'} \right)^\prime } = {\left[ {\frac{1}{4}{{\left( {x + 1} \right)}^{ - \large\frac{3}{4}\normalsize}}} \right]^\prime } } = {\frac{1}{4} \cdot \left( { - \frac{3}{4}} \right){\left( {x + 1} \right)^{ - \large\frac{7}{4}\normalsize}} } = { - \frac{3}{{16}} \cdot \frac{1}{{{{\left( {x + 1} \right)}^{\large\frac{7}{4}\normalsize}}}} } = { - \frac{3}{{16\sqrt[\large 4\normalsize]{{{{\left( {x + 1} \right)}^7}}}}}.} \]
   Example 3
Calculate \(y''\) for the parabola equation \({y^2} = 4x.\)

Solution.
By implicit differentiation, \[2yy' = 4,\;\; \Rightarrow yy' = 2.\] Differentiating again and using the product rule, we obtain \[y'y' + yy'' = 0.\] Multiply both sides by \({y^2}\): \[{y^2}{\left( {y'} \right)^2} + {y^3}y'' = 0.\] Since \(yy' = 2\) and, hence, \({\left( {yy'} \right)^2} = 4,\), we can write the last equation as \[4 + {y^3}y'' = 0.\] Then \[y'' = - \frac{4}{{{y^3}}}.\]
   Example 4
Given the function \(y = {\left( {2x + 1} \right)^3}\left( {x - 1} \right).\) Find all derivatives of the \(n\)th order from \(n = 1\) to \(n = 5.\)

Solution.
First we convert the given function into a polynomial: \[ {y = {\left( {2x + 1} \right)^3}\left( {x - 1} \right) } = {\left[ {{{\left( {2x} \right)}^3} + 3 \cdot {{\left( {2x} \right)}^2} \cdot 1 + 3 \cdot 2x \cdot {1^2} + {1^3}} \right]\left( {x - 1} \right) } = {\left( {8{x^3} + 12{x^2} + 6x + 1} \right)\left( {x - 1} \right) } = {\color{blue}{8{x^4}} + \color{red}{12{x^3}} + \color{maroon}{6{x^2}} + \color{green}x - \color{red}{8{x^3}} - \color{maroon}{12{x^2}} - \color{green}{6x} - \color{coral}1 } = {\color{blue}{8{x^4}} + \color{red}{4{x^3}} - \color{maroon}{6{x^2}} - \color{green}{5x} - \color{coral}1.} \] Now we successively calculate the derivatives from \(1\)st to \(5\)th order: \[ {y' = {\left( {8{x^4} + 4{x^3} - 6{x^2} - 5x - 1} \right)^\prime } = 32{x^3} + 12{x^2} - 12x - 5,}\;\;\; {y'' = {\left( {y'} \right)^\prime } = {\left( {32{x^3} + 12{x^2} - 12x - 5} \right)^\prime } = 96{x^2} + 24x - 12,}\;\;\; {y''' = {\left( {y''} \right)^\prime } = {\left( {96{x^2} + 24x - 12} \right)^\prime } = 192x + 24,}\;\;\; {{y^{\left( 4 \right)}} = {\left( {y'''} \right)^\prime } = {\left( {192x + 24} \right)^\prime } = 192,}\;\;\; {{y^{\left( 5 \right)}} = {\left( {{y^{\left( 4 \right)}}} \right)^\prime } = {\left( {192} \right)^\prime } = 0.} \]
   Example 5
Find the \(n\)th order derivative of the natural logarithm function \(y = \ln x.\)

Solution.
We calculate several successive derivatives of the given function: \[ {y' = {\left( {\ln x} \right)^\prime } = \frac{1}{x},}\;\;\; {y'' = {\left( {y'} \right)^\prime } = {\left( {\frac{1}{x}} \right)^\prime } = {\left( {{x^{ - 1}}} \right)^\prime } = - {x^{ - 2}} = - \frac{1}{{{x^2}}},}\;\;\; {y''' = {\left( {y''} \right)^\prime } = {\left( { - \frac{1}{{{x^2}}}} \right)^\prime } = 2{x^{ - 3}} = \frac{2}{{{x^3}}},}\;\;\; {{y^{\left( 4 \right)}} = {\left( {y'''} \right)^\prime } = {\left( {\frac{2}{{{x^3}}}} \right)^\prime } = - 6{x^{ - 4}} = - \frac{6}{{{x^4}}},}\;\;\; {{y^{\left( 5 \right)}} = {\left( {{y^{\left( 4 \right)}}} \right)^\prime } = {\left( { - \frac{6}{{{x^4}}}} \right)^\prime } = 24{x^{ - 5}} = \frac{{24}}{{{x^5}}}.} \] We see that the derivative of an arbitrary \(n\)th order is given by \[{y^{\left( n \right)}} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!}}{{{x^n}}}.\] A rigorous justification of this formula can be obtained using the method of mathematical induction.

   Example 6
Find all derivatives of the sine function.

Solution.
Calculate a few derivatives starting from the first one: \[ {y' = {\left( {\sin x} \right)^\prime } = \cos x = \sin \left( {x + \frac{\pi }{2}} \right),}\;\;\; {y'' = {\left( {\cos x} \right)^\prime } = - \sin x = \sin \left( {x + 2 \cdot \frac{\pi }{2}} \right),}\;\;\; {y''' = {\left( { - \sin x} \right)^\prime } = - \cos x = \sin \left( {x + 3 \cdot \frac{\pi }{2}} \right),}\;\;\; {{y^{IV}} = {\left( { - \cos x} \right)^\prime } = \sin x = \sin \left( {x + 4 \cdot \frac{\pi }{2}} \right).} \] Obviously, the \(n\)-order derivative is expressed by the formula \[{y^{\left( n \right)}} = {\left( {\sin x} \right)^{\left( n \right)}} = \sin \left( {x + \frac{{n\pi }}{2}} \right).\] A rigorous proof of this formula can be done by induction.

   Example 7
Find all derivatives of the cosine function.

Solution.
Analogously to Example \(6,\) we find the first few derivatives of the cosine function: \[ {y' = {\left( {\cos x} \right)^\prime } = - \sin x = \cos \left( {x + \frac{\pi }{2}} \right),}\;\;\; {y'' = {\left( { - \sin x} \right)^\prime } = - \cos x = \cos \left( {x + 2 \cdot \frac{\pi }{2}} \right),}\;\;\; {y''' = {\left( { - \cos x} \right)^\prime } = \sin x = \cos \left( {x + 3 \cdot \frac{\pi }{2}} \right),}\;\;\; {{y^{IV}} = {\left( {\sin x} \right)^\prime } = \cos x = \cos \left( {x + 4 \cdot \frac{\pi }{2}} \right).} \] Clearly that the next \(5\)th order derivative coincides with the first derivaive, the \(6\)th with the \(2\)nd and so on. Thus, the \(n\)th order derivative of the cosine function is described by the formula \[{y^{\left( n \right)}} = {\left( {\cos x} \right)^{\left( n \right)}} = \cos \left( {x + \frac{{n\pi }}{2}} \right).\]
   Example 8
Find all derivatives of the function \(y = {\large\frac{1}{x}\normalsize}.\)

Solution.
We find a few first derivatives: \[ {y' = {\left( {\frac{1}{x}} \right)^\prime } = - \frac{1}{{{x^2}}},}\;\; {{y'' = {\left( { - \frac{1}{{{x^2}}}} \right)^\prime } = - 1 \cdot {\left( {{x^{ - 2}}} \right)^\prime }} = {- 1 \cdot \left( { - 2} \right) \cdot {x^{ - 3}} = \frac{{{{\left( { - 1} \right)}^2} \cdot 2}}{{{x^3}}},}}\;\; {{y''' = {\left( {\frac{{{{\left( { - 1} \right)}^2} \cdot 2}}{{{x^3}}}} \right)^\prime } = {\left( { - 1} \right)^2} \cdot 2 \cdot {\left( {{x^{ - 3}}} \right)^\prime } } = {{\left( { - 1} \right)^3} \cdot 2 \cdot 3 \cdot {x^{ - 4}} = \frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 3}}{{{x^4}}},}}\;\; {{{y^{IV}} = {\left( {\frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 3}}{{{x^4}}}} \right)^\prime } = {\left( { - 1} \right)^3} \cdot 2 \cdot 3 \cdot {\left( {{x^{ - 4}}} \right)^\prime } } = {{\left( { - 1} \right)^4} \cdot 4! \cdot {x^{ - 5}} = \frac{{{{\left( { - 1} \right)}^4}4!}}{{{x^5}}}.}} \] This is enough to detect the general pattern: \[{y^{\left( n \right)}} = \frac{{{{\left( { - 1} \right)}^n}n!}}{{{x^{n + 1}}}}.\]
   Example 9
Determine the second derivative of the function \(y = \arcsin {\large\frac{{{x^2} - 1}}{{{x^2} + 1}}\normalsize}.\)

Solution.
Differentiating as a composite function, we find the first derivative: \[\require{cancel} {y = {\left( {\arcsin \frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)^\prime } } = {\frac{1}{{\sqrt {1 - {{\left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)}^2}} }} \cdot {\left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)^\prime } } = {\frac{1}{{\sqrt {\frac{{{{\left( {{x^2} + 1} \right)}^2} - {{\left( {{x^2} - 1} \right)}^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}} }} \cdot \frac{{2x \cdot \left( {{x^2} + 1} \right) - \left( {{x^2} - 1} \right) \cdot 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} } = {\frac{{{x^2} + 1}}{{\sqrt {\cancel{\color{blue}{x^4}} + \color{red}{2{x^2}} + \cancel{\color{green}1} - \cancel{\color{blue}{x^4}} + \color{red}{2{x^2}} - \cancel{\color{green}1}} }} \cdot \frac{{\cancel{\color{maroon}{2{x^3}}} + \color{darkmagenta}{2x} - \cancel{\color{maroon}{2{x^3}}} + \color{darkmagenta}{2x}}}{{{{\left( {{x^2} + 1} \right)}^2}}} } = {\frac{{\color{darkmagenta}{4x}}}{{\sqrt {\color{red}{4{x^2}}} \left( {{x^2} + 1} \right)}} } = {\frac{{4x}}{{2\left| x \right|\left( {{x^2} + 1} \right)}} } = {\frac{{2x}}{{\left| x \right|\left( {{x^2} + 1} \right)}}.} \] Here we can represent \({\left| x \right|}\) as \(x\,\text{sign}\,x,\) where \[ \text{sign}\,x = \begin{cases} - 1, & \text{if}\;\;x < 0 \\ 0, & \text{if} \;\;x = 0 \\ + 1, & \text{if} \;\;x > 0 \end{cases} .\] Then \[ {y' = \frac{{2x}}{{\left| x \right|\left( {{x^2} + 1} \right)}} } = {\frac{{2\cancel{x}}}{{\cancel{x} \,\text{sign}\,x\left( {{x^2} + 1} \right)}} } = {\frac{{2\,\text{sign}\,x}}{{{x^2} + 1}}.} \] We now calculate the second derivative differentiating the previous expression as a quotient of two functions: \[ {y'' = {\left( {\frac{{2\,\text{sign}\,x}}{{{x^2} + 1}}} \right)^\prime } } = {\frac{{{{\left( {2\,\text{sign}\,x} \right)}^\prime }\left( {{x^2} + 1} \right) } = {2\,\text{sign}\,x{{\left( {{x^2} + 1} \right)}^\prime }}}{{{{\left( {{x^2} + 1} \right)}^2}}} } = {\frac{{0 \cdot \left( {{x^2} + 1} \right) - 2\,\text{sign}\,x \cdot 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} } = { - \frac{{4x\,\text{sign}\,x}}{{{{\left( {{x^2} + 1} \right)}^2}}}.} \]
   Example 10
Find the third derivative of the function \(y = {\large\frac{{{x^3}}}{{x - 1}}\normalsize}.\)

Solution.
Differentiate successively the given function: \[ {y' = {\left( {\frac{{{x^3}}}{{x - 1}}} \right)^\prime } } = {\frac{{{{\left( {{x^3}} \right)}^\prime }\left( {x - 1} \right) - {x^3}{{\left( {x - 1} \right)}^\prime }}}{{{{\left( {x - 1} \right)}^2}}} } = {\frac{{3{x^2}\left( {x - 1} \right) - {x^3} \cdot 1}}{{{{\left( {x - 1} \right)}^2}}} } = {\frac{{\color{blue}{3{x^3}} - \color{red}{3{x^2}} - \color{blue}{x^3}}}{{{{\left( {x - 1} \right)}^2}}} } = {\frac{\color{blue}{{2{x^3}} - \color{red}{3{x^2}}}}{{{{\left( {x - 1} \right)}^2}}},} \] \[ {y'' = {\left( {y'} \right)^\prime } = {\left( {\frac{{2{x^3} - 3{x^2}}}{{{{\left( {x - 1} \right)}^2}}}} \right)^\prime } } = {\frac{{{{\left( {2{x^3} - 3{x^2}} \right)}^\prime }{{\left( {x - 1} \right)}^2} - \left( {2{x^3} - 3{x^2}} \right){{\left( {{{\left( {x - 1} \right)}^2}} \right)}^\prime }}}{{{{\left( {x - 1} \right)}^4}}} } = {\frac{{\left( {6{x^2} - 6x} \right){{\left( {x - 1} \right)}^2} - \left( {2{x^3} - 3{x^2}} \right) \cdot 2\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^4}}} } = {\frac{{6x\left( {{x^2} - 2x + 1} \right) - 4{x^3} + 6{x^2}}}{{{{\left( {x - 1} \right)}^3}}} } = {\frac{{\color{blue}{6{x^3}} - \color{red}{12{x^2}} + \color{green}{6x} - \color{blue}{4{x^3}} + \color{red}{6{x^2}}}}{{{{\left( {x - 1} \right)}^3}}} } = {\frac{{\color{blue}{2{x^3}} - \color{red}{6{x^2}} + \color{green}{6x}}}{{{{\left( {x - 1} \right)}^3}}}.} \] \[ {y''' = {\left( {y''} \right)^\prime } = {\left( {\frac{{2{x^3} - 6{x^2} + 6x}}{{{{\left( {x - 1} \right)}^3}}}} \right)^\prime } } = {\frac{{{{\left( {2{x^3} - 6{x^2} + 6x} \right)}^\prime }{{\left( {x - 1} \right)}^3} - \left( {2{x^3} - 6{x^2} + 6x} \right){{\left( {{{\left( {x - 1} \right)}^3}} \right)}^\prime }}}{{{{\left( {x - 1} \right)}^6}}} } = {\frac{{\left( {6{x^2} - 12x + 6} \right){{\left( {x - 1} \right)}^3} - \left( {2{x^3} - 6{x^2} + 6x} \right) \cdot 3{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^6}}} } = {\frac{{\left[ {\left( {6{x^2} - 12x + 6} \right)\left( {x - 1} \right) - 3\left( {2{x^3} - 6{x^2} + 6x} \right)} \right]{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^6}}} } = {\frac{{\cancel{\color{blue}{6{x^3}}} - \cancel{\color{red}{12{x^2}}} + \cancel{\color{green}{6x}} - \cancel{\color{red}{6{x^2}}} + \cancel{\color{green}{12x}} - \color{maroon}{6} - \cancel{\color{blue}{6{x^3}}} + \cancel{\color{red}{18{x^2}}} - \cancel{\color{green}{18x}}}}{{{{\left( {x - 1} \right)}^4}}} } = {- \frac{\color{maroon}{6}}{{{{\left( {x - 1} \right)}^4}}}. } \]
   Example 11
Find the second derivative of the implicitly defined function \({x^2} + {y^2} = {R^2}\) (canonical equation of a circle).

Solution.
Given that \(y\) is a function of \(x\) and differentiating both sides of the equation, we find the first derivative: \[ {{\left( {{x^2} + {y^2}} \right)^\prime } = {\left( {{R^2}} \right)^\prime },}\;\; {\Rightarrow 2x + 2yy' = 0,}\;\; {\Rightarrow yy' = - x,}\;\; {\Rightarrow y' = - \frac{x}{y}.} \] We differentiate this expression again: \[ {{\left( {y'} \right)^\prime } = {\left( { - \frac{x}{y}} \right)^\prime },}\;\; {\Rightarrow y'' = - \frac{{x'y - xy'}}{{{y^2}}},}\;\; {\Rightarrow y'' = - \frac{{y - xy'}}{{{y^2}}} = \frac{{xy' - y}}{{{y^2}}}.} \] Substituting the first derivative \(y'\) into this formula, we have \[ {y'' = \frac{{xy' - y}}{{{y^2}}} } = {\frac{{x\left( { - \frac{x}{y}} \right) - y}}{{{y^2}}} } = {\frac{{ - \frac{{{x^2}}}{y} - y}}{{{y^2}}} } = {\frac{{ - {x^2} - {y^2}}}{{{y^3}}} } = { - \frac{{{x^2} + {y^2}}}{{{y^3}}} } = { - \frac{{{R^2}}}{{{y^3}}}.} \]
   Example 12
Find the \(n\)th order derivative of the function \(y = {3^{2x + 1}}.\)

Solution.
We calculate successively several derivatives, starting from the first one. \[ {y' = {\left( {{3^{2x + 1}}} \right)^\prime } } = {{3^{2x + 1}} \cdot \ln 3 \cdot {\left( {2x + 1} \right)^\prime } } = {{3^{2x + 1}} \cdot 2\ln 3,} \] \[ {y'' = {\left( {y'} \right)^\prime } } = {{\left( {{3^{2x + 1}} \cdot 2\ln 3} \right)^\prime } } = {{\left( {{3^{2x + 1}}} \right)^\prime } \cdot 2\ln 3 } = {{3^{2x + 1}} \cdot {2^2}{\ln ^2}3,} \] \[ {y''' = {\left( {y''} \right)^\prime } } = {{\left( {{3^{2x + 1}} \cdot {2^2}{{\ln }^2}3} \right)^\prime } } = {{\left( {{3^{2x + 1}}} \right)^\prime } \cdot {2^2}{\ln ^2}3 } = {{3^{2x + 1}} \cdot {2^3}{\ln ^3}3.} \] It follows from here that the \(n\)th order derivative is given by \[ {{y^{\left( n \right)}} = {\left( {{3^{2x + 1}}} \right)^{\left( n \right)}} } = {{3^{2x + 1}} \cdot {2^n}{\ln ^n}3.} \] A rigorous proof can be carried out by induction. Clearly that this formula is valid for \(n = 1\). Suppose that it is true \(n = k:\) \[ {{y^{\left( k \right)}} = {\left( {{3^{2x + 1}}} \right)^{\left( k \right)}} } = {{3^{2x + 1}} \cdot {2^k}\,{\ln ^k}3.} \] Then for \(n = k + 1\) we have \[ {{y^{\left( {k + 1} \right)}} = {\left( {{3^{2x + 1}}} \right)^{\left( {k + 1} \right)}} } = {{\left[ {{{\left( {{3^{2x + 1}}} \right)}^{\left( k \right)}}} \right]^\prime } } = {{\left( {{3^{2x + 1}} \cdot {2^k}\,{{\ln }^k}3} \right)^\prime } } = {{\left( {{3^{2x + 1}}} \right)^\prime } \cdot {2^k}\,{\ln ^k}3 } = {{3^{2x + 1}} \cdot {2^{k + 1}}{\ln ^{k + 1}}3,} \] i.e. this formula holds for \(n = k + 1.\) Consequently, it is true for any natural number \(n.\)

   Example 13
Find the \(n\)th order derivative of the power function \(y = {x^m}\) where \(m\) is a real number.

Solution.
We calculate several first derivatives: \[ {y' = {\left( {{x^m}} \right)^\prime } = m{x^{m - 1}},}\;\; {y'' = {\left( {y'} \right)^\prime } = {\left( {m{x^{m - 1}}} \right)^\prime } = {m\left( {m - 1} \right){x^{m - 2}},}}\;\; {y''' = {\left( {y''} \right)^\prime } = {\left[ {m\left( {m - 1} \right){x^{m - 2}}} \right]^\prime } = {\left[ {m\left( {m - 1} \right)\left( {m - 2} \right)} \right]{x^{m - 3}}.}} \] Hence it is easy to establish a general expression for the \(n\)th order derivative: \[{y^{\left( n \right)}} = m\left( {m - 1} \right)\left( {m - 2} \right) \ldots \left( {m - n + 1} \right){x^{m - n}}.\] We prove this by induction. Obviously, this formula is valid for \(n = 1\). Assuming that it is true for the degree \(n\), we differentiate it again and find the derivative of \(\left( {n + 1} \right)\)th order: \[ {{y^{\left( {n + 1} \right)}} = {\left[ {{y^{\left( n \right)}}} \right]^\prime } } = {{\left[ {m\left( {m - 1} \right)\left( {m - 2} \right) \ldots \left( {m - n + 1} \right){x^{m - n}}} \right]^\prime } } = {m\left( {m - 1} \right)\left( {m - 2} \right) \ldots \left( {m - n + 1} \right){\left( {{x^{m - n}}} \right)^\prime } } = {m\left( {m - 1} \right)\left( {m - 2} \right) \ldots \left( {m - n + 1} \right){x^{m - \left( {n + 1} \right)}}.} \] As can be seen, the derivative of \(\left( {n + 1} \right)\)th order is expressed by the same formula as the \(n\)th order derivative (only the number \(n\) is replaced by \(n + 1\)). Consequently, the resulting expression is valid for any positive integer value of \(n\) (\(n\) is the order of the derivative).

Note that the exponent \(m\), generally speaking, is a real number. If we consider only the natural values of \(m\), then the formula for the derivative can be written in a more compact form: \[{y^{\left( n \right)}} = {\left( {{x^m}} \right)^{\left( n \right)}} = \frac{{m!}}{{\left( {m - n} \right)!}}{x^{m - n}},\] where \(n \le m\). All other derivatives of order \(n > m\) are equal to zero.

   Example 14
Find the \(n\)th order derivative of the square root \(y = \sqrt x .\)

Solution.
We use the result of Example \(13\), where the \(n\)th order derivative of the power function with an arbitrary real exponent \(m\) is derived. In this case we have \[y = \sqrt x = {x^{\large\frac{1}{2}\normalsize}}\;\;\left( {m = \frac{1}{2}} \right).\] Then the derivative is written as \[ {y = {\left( {{x^{\large\frac{1}{2}\normalsize}}} \right)^\prime } } = {\frac{1}{2} \cdot \left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right) \cdots \left( {\frac{1}{2} - n + 1} \right){x^{\large\frac{1}{2}\normalsize - n}} } = {\frac{1}{2} \cdot \left( { - \frac{1}{2}} \right) \cdot \left( { - \frac{3}{2}} \right) \cdot \left( { - \frac{5}{2}} \right) \cdots \left[ { - \left( {n - \frac{3}{2}} \right)} \right]\frac{{{x^{\large\frac{1}{2}\normalsize}}}}{{{x^n}}} } = {\frac{1}{2} \cdot {\left( { - 1} \right)^{n - 1}}\left[ {\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2} \cdots \frac{{2n - 3}}{2}} \right]\frac{{\sqrt x }}{{{x^n}}} } = {\frac{1}{2} \cdot {\left( { - 1} \right)^{n - 1}} \cdot \frac{1}{{{2^{n - 1}}}} \cdot \left[ {1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right)} \right]\frac{{\sqrt x }}{{{x^n}}} } = {\frac{{{{\left( { - 1} \right)}^{n - 1}}\left[ {1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right)} \right]\sqrt x }}{{{x^n}}}.} \] For \(n = 1\), the derivative is \[ {y' = \frac{{{{\left( { - 1} \right)}^0} \cdot 1 \cdot \sqrt x }}{{{{\left( {2x} \right)}^1}}} } = {\frac{{\sqrt x }}{{2x}} } = {\frac{1}{{2\sqrt x }}.} \] Provided \(n \ge 2\), the product of odd numbers in square brackets can be written in terms of the double factorial: \[1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right) = \left( {2n - 3} \right)!!\] Hence for \(n \ge 2\), the \(n\)th order derivative is expressed by the general formula \[ {{y^{\left( n \right)}} = \frac{1}{2} \cdot {\left( { - 1} \right)^{n - 1}} \cdot \frac{1}{{{2^{n - 1}}}} \cdot \left( {2n - 3} \right)!!\frac{{\sqrt x }}{{{x^n}}} } = {\frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {2n - 3} \right)!!\sqrt x }}{{{{\left( {2x} \right)}^n}}}.} \] In particular, \[ {y'' = \frac{{{{\left( { - 1} \right)}^1}1!!\sqrt x }}{{{{\left( {2x} \right)}^2}}} = - \frac{{\sqrt x }}{{4{x^2}}} = - \frac{1}{{4\sqrt {{x^3}} }},}\;\; {y''' = \frac{{{{\left( { - 1} \right)}^2}3!!\sqrt x }}{{{{\left( {2x} \right)}^3}}} = \frac{{3\sqrt x }}{{8{x^3}}} = \frac{3}{{8\sqrt {{x^5}} }}.} \]
   Example 15
Find the \(n\)th order derivative of the cube root \(y = \sqrt[\large 3\normalsize]{x}.\)

Solution.
The first derivative of the cube root is given by \[ {y' = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } = {\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{\large\frac{1}{3}\normalsize - 1}} } = {\frac{1}{3}{x^{ - \large\frac{2}{3}\normalsize}} } = {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}}.} \] Next, we use the general formula for the \(n\)th order derivative of the power function \(y = {x^m}\) (Example \(13\)): \[{\left( {{x^m}} \right)^{\left( n \right)}} = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n + 1} \right){x^{m - n}}.\] In our case \(m = {\large\frac{1}{3}\normalsize}.\) Consequently, the derivative is as follows: \[ {{y^{\left( n \right)}} = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^{\left( n \right)}} } = {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^{\left( n \right)}} } = {\frac{1}{3}\left( {\frac{1}{3} - 1} \right)\left( {\frac{1}{3} - 2} \right)\left( {\frac{1}{3} - 3} \right) \cdots \left( {\frac{1}{3} - n + 1} \right){x^{\large\frac{1}{3}\normalsize - n}} } = {\frac{1}{3} \cdot \left( { - \frac{2}{3}} \right) \cdot \left( { - \frac{5}{3}} \right) \cdot \left( { - \frac{8}{3}} \right) \cdots \left[ { - \left( {n - \frac{4}{3}} \right)} \right]\frac{{{x^{\large\frac{1}{3}\normalsize}}}}{{{x^n}}} } = {\frac{1}{3} \cdot {\left( { - 1} \right)^{n - 1}}\left[ {\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdots \frac{{3n - 4}}{3}} \right]\frac{{\sqrt[\large 3\normalsize]{x}}}{{{x^n}}} } = {\frac{{{{\left( { - 1} \right)}^{n - 1}}\left[ {2 \cdot 5 \cdot 8 \cdots \left( {3n - 4} \right)} \right]\sqrt[\large 3\normalsize]{x}}}{{{{\left( {3x} \right)}^n}}}.} \] where \(n \ge 2.\) Specifically, the second and third derivatives of the cube root are expressed by the formulas \[ {y'' = \frac{{{{\left( { - 1} \right)}^1} \cdot 2 \cdot \sqrt[\large 3\normalsize]{x}}}{{{{\left( {3x} \right)}^2}}} } = { - \frac{{2\sqrt[\large 3\normalsize]{x}}}{{9{x^2}}} } = {- \frac{2}{{9\sqrt[3]{{{x^5}}}}},}\;\; {y''' = \frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 5 \cdot \sqrt[\large 3\normalsize]{x}}}{{{{\left( {3x} \right)}^3}}} } = {\frac{{10\sqrt[\large 3\normalsize]{x}}}{{27{x^3}}} } = {\frac{{10}}{{27\sqrt[\large 3\normalsize]{{{x^8}}}}}.} \]
   Example 16
Find the first and second derivatives with respect to \(x\) of the function given parametrically: \[ \left\{ \begin{aligned} x\left( t \right) &= {e^t}\sin t \\ y\left( t \right) &= {e^t}\cos t \end{aligned} \right.. \]

Solution.
Compute the first derivative \({y'_x}\) by the formula: \[y' = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}}.\] Consequently, \[ {y' = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {{e^t}\cos t} \right)}^\prime }}}{{{{\left( {{e^t}\sin t} \right)}^\prime }}} } = {\frac{{{{\left( {{e^t}} \right)}^\prime }\cos t + {e^t}{{\left( {\cos t} \right)}^\prime }}}{{{{\left( {{e^t}} \right)}^\prime }\sin t + {e^t}{{\left( {\sin t} \right)}^\prime }}} } = {\frac{{{e^t}\cos t - {e^t}\sin t}}{{{e^t}\sin t + {e^t}\cos t}} } = {\frac{{\cancel{e^t}\left( {\cos t - \sin t} \right)}}{{\cancel{e^t}\left( {\sin t + \cos t} \right)}} } = {\frac{{\cos t - \sin t}}{{\cos t + \sin t}}.} \] Differentiating again, we find the second derivative with respect to \(x:\) \[ {y'' = {y''_{xx}} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} } = {\frac{{{{\left( {\frac{{\cos t - \sin t}}{{\cos t + \sin t}}} \right)}^\prime }}}{{{{\left( {{e^t}\sin t} \right)}^\prime }}}.} \] Calculate separately the derivative in the numerator: \[ {{\left( {{y'_x}} \right)'_t} = {\left( {\frac{{\cos t - \sin t}}{{\cos t + \sin t}}} \right)^\prime } } = {\frac{{{{\left( {\cos t - \sin t} \right)}^\prime }\left( {\cos t + \sin t} \right) - \left( {\cos t - \sin t} \right){{\left( {\cos t + \sin t} \right)}^\prime }}}{{{{\left( {\cos t + \sin t} \right)}^2}}} } = {\frac{{\left( { - \sin t - \cos t} \right)\left( {\cos t + \sin t} \right) - \left( {\cos t - \sin t} \right)\left( { - \sin t + \cos t} \right)}}{{{{\left( {\cos t + \sin t} \right)}^2}}} } = {\frac{{ - {{\left( {\cos t + \sin t} \right)}^2} - {{\left( {\cos t - \sin t} \right)}^2}}}{{{{\left( {\cos t + \sin t} \right)}^2}}} } = {\frac{{{{\cos }^2}t + \cancel{2\sin t\cos t} + {{\sin }^2}t + {{\cos }^2}t - \cancel{2\sin t\cos t} + {{\sin }^2}t}}{{{{\left( {\cos t + \sin t} \right)}^2}}} } = { - \frac{2}{{{{\left( {\cos t + \sin t} \right)}^2}}}.} \] Since the derivative in the denominator is equal to \[{x'_t} = {e^t}\left( {\sin t + \cos t} \right),\] we obtain the following expression for the second derivative of the original function: \[ {y'' = {y''_{xx}} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} } = {\frac{{ - \frac{2}{{{{\left( {\cos t + \sin t} \right)}^2}}}}}{{{e^t}\left( {\sin t + \cos t} \right)}} } = { - \frac{{2{e^{ - t}}}}{{{{\left( {\cos t + \sin t} \right)}^3}}}.} \]
   Example 17
Given the equation of an ellipse in parametric form: \[x = a\cos t,\;\;y = b\sin t,\] where \(a\), \(b\) are semi-axes of the ellipse, \(t\) is a parameter. Find the first, second and third derivatives of the function \(y\) with respect to \(x.\)

Solution.
Differentiating the given parametric function successively, we obtain: \[ {y = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} } = {\frac{{{{\left( {b\sin t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} } = {\frac{{b\cos t}}{{ - a\sin t}} } = { - \frac{b}{a}\cot t,} \] \[ {y'' = {y''_{xx}} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} } = {\frac{{{{\left( { - \frac{b}{a}\cot t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} } = {\frac{{\left( { - \frac{b}{a}} \right)\left( { - \frac{1}{{{{\sin }^2}t}}} \right)}}{{\left( { - a\sin t} \right)}} } = { - \frac{b}{{{a^2}}}\frac{1}{{{{\sin }^3}t}} } = { - \frac{b}{{{a^2}}}{\csc ^3}t,} \] \[ {y''' = {y'''_{xxx}} = \frac{{{{\left( {{y''_{xx}}} \right)}'_t} }}{{{x'_t}}} } = {\frac{{{{\left( { - \frac{b}{{{a^2}}}{{\csc }^3}t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} } = {\frac{{\left( { - \frac{b}{{{a^2}}}} \right) \cdot 3{{\csc }^2}t \cdot {{\left( {\csc t} \right)}^\prime }}}{{\left( { - a\sin t} \right)}} } = { - \frac{{3b}}{{{a^3}}} \cdot \frac{{{{\csc }^2}t \cdot \left( { - \cot t} \right) \cdot \csc t}}{{\left( { - \sin t} \right)}} } = { - \frac{{3b}}{{{a^3}}} \cdot \frac{{{{\csc }^3}t\cot t}}{{\sin t}} } = { - \frac{{3b}}{{{a^3}}}{\csc ^4}t\cot t.} \]
   Example 18
Find the third derivative of the function given by the equation \({x^2} + 3xy + {y^2} = 1.\)

Solution.
One time differentiation with respect to \(x\) leads to the following expression: \[ {{\left( {{x^2} + 3xy + {y^2}} \right)^\prime } = {1^\prime },}\;\; {\Rightarrow 2x + 3\left( {x'y + xy'} \right) + 2yy' = 0,}\;\; {\Rightarrow 2x + 3y + 3xy' + 2yy' = 0,}\;\; {\Rightarrow 2x + 3y + \left( {3x + 2y} \right)y' = 0,}\;\; {\Rightarrow y' = - \frac{{2x + 3y}}{{3x + 2y}}.} \] We differentiate the last expression again considering \(y\) as a composite function: \[ {y'' = {\left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)^\prime } } = { - \frac{{{{\left( {2x + 3y} \right)}^\prime }\left( {3x + 2y} \right) - \left( {2x + 3y} \right){{\left( {3x + 2y} \right)}^\prime }}}{{{{\left( {3x + 2y} \right)}^2}}} } = { - \frac{{\left( {2 + 3y'} \right)\left( {3x + 2y} \right) - \left( {2x + 3y} \right)\left( {3 + 2y'} \right)}}{{{{\left( {3x + 2y} \right)}^2}}} } = { - \frac{{\cancel{\color{blue}{6x}} + \color{green}{9xy'} + \color{red}{4y} + \cancel{\color{maroon}{6yy'}} - \cancel{\color{blue}{6x}} - \color{red}{9y} - \color{green}{4xy'} - \cancel{\color{maroon}{6yy'}}}}{{{{\left( {3x + 2y} \right)}^2}}} } = { - \frac{{\color{green}{5xy'} - \color{red}{5y}}}{{{{\left( {3x + 2y} \right)}^2}}} } = {\frac{{5y - 5xy'}}{{{{\left( {3x + 2y} \right)}^2}}}.} \] Substitute the explicit expression for the first derivative \(y':\) \[ {y'' = \frac{{5y - 5xy'}}{{{{\left( {3x + 2y} \right)}^2}}} } = {\frac{{5y - 5x\left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)}}{{{{\left( {3x + 2y} \right)}^2}}} } = {\frac{{5y\left( {3x + 2y} \right) + 5x\left( {2x + 3y} \right)}}{{{{\left( {3x + 2y} \right)}^3}}} } = {\frac{{15xy + 10{y^2} + 10{x^2} + 15xy}}{{{{\left( {3x + 2y} \right)}^3}}} } = {\frac{{10\left( {{x^2} + 3xy + {y^2}} \right)}}{{{{\left( {3x + 2y} \right)}^3}}}.} \] Since \({x^2} + 3xy + {y^2} = 1,\) we obtain the following expression for the second derivative \(y'':\) \[y'' = \frac{{10}}{{{{\left( {3x + 2y} \right)}^3}}}.\] Similarly, differentiating once more, we get the third derivative: \[ {y''' = {\left[ {\frac{{10}}{{{{\left( {3x + 2y} \right)}^3}}}} \right]^\prime } } = {\frac{{{{\left( {10} \right)}^\prime } \cdot {{\left( {3x + 2y} \right)}^3} - 10 \cdot {{\left[ {{{\left( {3x + 2y} \right)}^3}} \right]}^\prime }}}{{{{\left( {3x + 2y} \right)}^6}}} } = {\frac{{0 \cdot {{\left( {3x + 2y} \right)}^3} - 10 \cdot 3{{\left( {3x + 2y} \right)}^2} \cdot {{\left( {3x + 2y} \right)}^\prime }}}{{{{\left( {3x + 2y} \right)}^6}}} } = { - \frac{{30{{\left( {3x + 2y} \right)}^2}\left( {3 + 2y'} \right)}}{{{{\left( {3x + 2y} \right)}^6}}} } = { - \frac{{30\left( {3 + 2y'} \right)}}{{{{\left( {3x + 2y} \right)}^4}}} } = { - \frac{{90 + 60y'}}{{{{\left( {3x + 2y} \right)}^4}}}.} \] We substitute again the first derivative and find: \[ {y''' = - \frac{{90 + 60y'}}{{{{\left( {3x + 2y} \right)}^4}}} } = { - \frac{{90 + 60 \cdot \left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)}}{{{{\left( {3x + 2y} \right)}^4}}} } = { - \frac{{90\left( {3x + 2y} \right) - 60\left( {2x + 3y} \right)}}{{{{\left( {3x + 2y} \right)}^5}}} } = { - \frac{{150}}{{{{\left( {3x + 2y} \right)}^5}}}.} \]
   Example 19
Find the second derivative of the function given by the equation \(x + y = {e^{x - y}}.\)

Solution.
Differentiating both sides in \(x\), we obtain: \[ {{\left( {x + y} \right)^\prime } = {\left( {{e^{x - y}}} \right)^\prime },}\;\; {\Rightarrow 1 + y' = {e^{x - y}} \cdot {\left( {x - y} \right)^\prime },}\;\; {\Rightarrow 1 + y' = {e^{x - y}}\left( {1 - y'} \right) = {e^{x - y}} - {e^{x - y}}y',}\;\; {\Rightarrow y' + {e^{x - y}}y' = {e^{x - y}} - 1,}\;\; {\Rightarrow y' = \frac{{{e^{x - y}} - 1}}{{{e^{x - y}} + 1}}.} \] Continuing the differentiation, we find the second derivative: \[ {y'' = {\left( {\frac{{{e^{x - y}} - 1}}{{{e^{x - y}} + 1}}} \right)^\prime } } = {\frac{{{{\left( {{e^{x - y}} - 1} \right)}^\prime }\left( {{e^{x - y}} + 1} \right) - \left( {{e^{x - y}} - 1} \right){{\left( {{e^{x - y}} + 1} \right)}^\prime }}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} } = {\frac{{{e^{x - y}}\left( {1 - y'} \right)\left( {{e^{x - y}} + 1} \right) - \left( {{e^{x - y}} - 1} \right){e^{x - y}}\left( {1 - y'} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} } = {\frac{{{e^{x - y}}\left( {1 - y'} \right)\left( {\cancel{e^{x - y}} + 1 - \cancel{e^{x - y}} + 1} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} } = {\frac{{2{e^{x - y}}\left( {1 - y'} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}}.} \] Substitute the expression for the first derivative: \[ {y'' = \frac{{2{e^{x - y}}\left( {1 - y'} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} } = {\frac{{2{e^{x - y}}\left( {1 - \frac{{{e^{x - y}} - 1}}{{{e^{x - y}} + 1}}} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} } = {\frac{{2{e^{x - y}} \cdot \frac{{\cancel{e^{x - y}} + 1 - \cancel{e^{x - y}} + 1}}{{{e^{x - y}} + 1}}}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} } = {\frac{{4{e^{x - y}}}}{{{{\left( {{e^{x - y}} + 1} \right)}^3}}}.} \] We now use the original equation, according to which \[{e^{x - y}} = x + y.\] As a result, we obtain the following expression for the derivative \(y'':\) \[ {y'' = \frac{{4{e^{x - y}}}}{{{{\left( {{e^{x - y}} + 1} \right)}^3}}} } = {\frac{{4\left( {x + y} \right)}}{{{{\left( {x + y + 1} \right)}^3}}}.} \]
   Example 20
A curve is described by the implicit equation \(xy = 2{x^3} - {y^2}\) and passes through the point \(\left( {1,1} \right).\) Find the first and second derivatives at this point.

Solution.
We calculate the first derivative: \[ {{\left( {xy} \right)^\prime } = {\left( {2{x^3} - {y^2}} \right)^\prime },}\;\; {\Rightarrow x'y + xy' = 6{x^2} - 2yy',}\;\; {\Rightarrow y + xy' = 6{x^2} - 2yy',}\;\; {\Rightarrow xy' + 2yy' = 6{x^2} - y,}\;\; {\Rightarrow y'\left( {x + 2y} \right) = 6{x^2} - y,}\;\; {\Rightarrow y' = \frac{{6{x^2} - y}}{{x + 2y}} } ={ \frac{{6 \cdot {1^2} - 1}}{{1 + 2 \cdot 1}} = \frac{5}{3}.} \] Differentiating again, we find the value of the second derivative at the given point: \[ {y'' = {\left( {\frac{{6{x^2} - y}}{{x + 2y}}} \right)^\prime } } = {\frac{{{{\left( {6{x^2} - y} \right)}^\prime }\left( {x + 2y} \right) - \left( {6{x^2} - y} \right){{\left( {x + 2y} \right)}^\prime }}}{{{{\left( {x + 2y} \right)}^2}}} } = {\frac{{\left( {12x - y'} \right)\left( {x + 2y} \right) - \left( {6{x^2} - y} \right)\left( {1 + 2y'} \right)}}{{{{\left( {x + 2y} \right)}^2}}} } = {\frac{{\color{blue}{12{x^2}} - xy' + 24xy - \cancel{\color{red}{2yy'}} - \color{blue}{6{x^2}} + y - 12{x^2}y' + \cancel{\color{red}{2yy'}}}}{{{{\left( {x + 2y} \right)}^2}}} } = {\frac{{\color{blue}{6{x^2}} - xy' + 24xy - 12{x^2}y' + y}}{{{{\left( {x + 2y} \right)}^2}}}.} \] Substitute the known values of \(x\), \(y\), \(y'\) to get: \[y'' = \frac{{6 \cdot {1^2} - 1 \cdot \frac{5}{3} + 24 \cdot 1 \cdot 1 - 12 \cdot {1^2} \cdot \frac{5}{3} + 1}}{{{{\left( {1 + 2 \cdot 1} \right)}^2}}} = \frac{{28}}{{27}}.\]

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