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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: $y'' = \frac{{6 \cdot {1^2} - 1 \cdot \frac{5}{3} + 24 \cdot 1 \cdot 1 - 12 \cdot {1^2} \cdot \frac{5}{3} + 1}}{{{{\left( {1 + 2 \cdot 1} \right)}^2}}} = \frac{{28}}{{27}}.$