Second Derivatives
Solved Problems
Example 7.
Find \(y^{\prime\prime},\) if \[y = x\sqrt {1 + {x^2}}.\]
Solution.
Take the first derivative using the product rule:
\[\require{cancel} y^\prime = \left( {x\sqrt {1 + {x^2}} } \right)^\prime = x^\prime\sqrt {1 + {x^2}} + x\left( {\sqrt {1 + {x^2}} } \right)^\prime = 1 \cdot \sqrt {1 + {x^2}} + x \cdot \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {1 + {x^2}} }} = \sqrt {1 + {x^2}} + \frac{{{x^2}}}{{\sqrt {1 + {x^2}} }} = \frac{{\color{blue}{1} + \color{red}{x^2} + \color{red}{x^2}}}{{\sqrt {1 + {x^2}} }} = \frac{{\color{blue}{1} + \color{red}{2{x^2}}}}{{\sqrt {1 + {x^2}} }}.\]
Differentiate this expression once more using the quotient rule:
\[y^{\prime\prime} = \left( {\frac{{1 + 2{x^2}}}{{\sqrt {1 + {x^2}} }}} \right)^\prime = \frac{{4x \cdot \sqrt {1 + {x^2}} - \left( {1 + 2{x^2}} \right) \cdot \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {1 + {x^2}} }}}}{{{{\left( {\sqrt {1 + {x^2}} } \right)}^2}}} = \frac{{4x\sqrt {1 + {x^2}} - \frac{{x\left( {1 + 2{x^2}} \right)}}{{\sqrt {1 + {x^2}} }}}}{{1 + {x^2}}} = \frac{{4x\left( {1 + {x^2}} \right) - x\left( {1 + 2{x^2}} \right)}}{{\sqrt {{{\left( {1 + {x^2}} \right)}^3}} }} = \frac{{\color{blue}{4x} + \color{red}{4{x^3}} - \color{blue}{x} - \color{red}{2{x^3}}}}{{\sqrt {{{\left( {1 + {x^2}} \right)}^3}} }} = \frac{{\color{red}{2{x^3}} + \color{blue}{3x}}}{{\sqrt {{{\left( {1 + {x^2}} \right)}^3}} }} = \frac{{x\left( {2{x^2} + 3} \right)}}{{\sqrt {{{\left( {1 + {x^2}} \right)}^3}} }}.\]
Example 8.
Find the second derivative of the function \[y = \sqrt[4]{{x + 1}}.\]
Solution.
We start with first derivative:
\[y' = \left( {\sqrt[4]{{x + 1}}} \right)^\prime = \left[ {{{\left( {x + 1} \right)}^{\frac{1}{4}}}} \right]^\prime = \frac{1}{4}{\left( {x + 1} \right)^{ - \frac{3}{4}}} = \frac{1}{{4\sqrt[4]{{{{\left( {x + 1} \right)}^3}}}}}.\]
Differentiate again to find the second derivative:
\[y^{\prime\prime = \left( {y'} \right)^\prime = \left[ {\frac{1}{4}{{\left( {x + 1} \right)}^{ - \frac{3}{4}}}} \right]^\prime } = \frac{1}{4} \cdot \left( { - \frac{3}{4}} \right){\left( {x + 1} \right)^{ - \frac{7}{4}}} = - \frac{3}{{16}} \cdot \frac{1}{{{{\left( {x + 1} \right)}^{\frac{7}{4}}}}} = - \frac{3}{{16\sqrt[4]{{{{\left( {x + 1} \right)}^7}}}}}.\]
Example 9.
Find \(y^{\prime\prime},\) if \[y = {x^x}.\]
Solution.
We use logarithmic differentiation to find the first derivative:
\[y = {x^x},\;\; \Rightarrow \ln y = \ln {x^x},\;\; \Rightarrow \ln y = x\ln x,\;\; \Rightarrow \left( {\ln y} \right)^\prime = \left( {x\ln x} \right)^\prime,\;\; \Rightarrow \frac{{y^\prime}}{y} = 1 \cdot \ln x + x \cdot \frac{1}{x},\;\; \Rightarrow y^\prime = y\left( {\ln x + 1} \right),\;\; \Rightarrow y^\prime = {x^x}\left( {\ln x + 1} \right).\]
Differentiating both sides once more, we obtain the second derivative:
\[y^\prime = y\left( {\ln x + 1} \right),\;\; \Rightarrow y^{\prime\prime} = \left[ {y\left( {\ln x + 1} \right)} \right]^\prime,\;\; \Rightarrow y^{\prime\prime} = y^\prime\left( {\ln x + 1} \right) + y\left( {\ln x + 1} \right)^\prime,\;\; \Rightarrow y^{\prime\prime} = {x^x}\left( {\ln x + 1} \right) \cdot \left( {\ln x + 1} \right) + {x^x} \cdot \frac{1}{x},\;\; \Rightarrow y^{\prime\prime} = {x^x}{\left( {\ln x + 1} \right)^2} + {x^{x - 1}}.\]
Example 10.
Calculate \(y^{\prime\prime}\) 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
\[y'y' + yy^{\prime\prime} = 0.\]
Multiply both sides by \({y^2}:\)
\[{y^2}{\left( {y'} \right)^2} + {y^3}y^{\prime\prime} = 0.\]
As \(yy' = 2\) and, hence, \({\left( {yy'} \right)^2} = 4,\) we can write the last equation as
\[4 + {y^3}y^{\prime\prime} = 0.\]
Then
\[y^{\prime\prime} = - \frac{4}{{{y^3}}}.\]
Example 11.
Determine the second derivative of the function \[y = \arcsin {\frac{{{x^2} - 1}}{{{x^2} + 1}}}.\]
Solution.
Differentiating as a composite function, we find the first derivative:
\[y = \left( {\arcsin \frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)^\prime = \frac{1}{{\sqrt {1 - {{\left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)}^2}} }} \cdot {\left( {\frac{{{x^2} - 1}}{{{x^2} + 1}}} \right)^\prime } = \frac{1}{{\sqrt {\frac{{{{\left( {{x^2} + 1} \right)}^2} - {{\left( {{x^2} - 1} \right)}^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}} }} \cdot \frac{{2x \cdot \left( {{x^2} + 1} \right) - \left( {{x^2} - 1} \right) \cdot 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \frac{{{4x}}}{{\sqrt {{4{x^2}}} \left( {{x^2} + 1} \right)}} = \frac{{4x}}{{2\left| x \right|\left( {{x^2} + 1} \right)}} = \frac{{2x}}{{\left| x \right|\left( {{x^2} + 1} \right)}}.\]
Here we can represent \({\left| x \right|}\) as \(x\,\text{sign}\,x,\) where
\[ \text{sign}\,x = \begin{cases} - 1, & \text{if}\;\;x \lt 0 \\ 0, & \text{if} \;\;x = 0 \\ + 1, & \text{if} \;\;x \gt 0 \end{cases} .\]
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:
\[y^{\prime\prime} = {\left( {\frac{{2\,\text{sign}\,x}}{{{x^2} + 1}}} \right)^\prime } = \frac{{0 \cdot \left( {{x^2} + 1} \right) - 2\,\text{sign}\,x \cdot 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}} = - \frac{{4x\,\text{sign}\,x}}{{{{\left( {{x^2} + 1} \right)}^2}}}.\]
Example 12.
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:
\[\left( {y'} \right)^\prime = \left( { - \frac{x}{y}} \right)^\prime,\;\; \Rightarrow y^{\prime\prime} = - \frac{{x'y - xy'}}{{{y^2}}},\;\; \Rightarrow y^{\prime\prime} = - \frac{{y - xy'}}{{{y^2}}} = \frac{{xy' - y}}{{{y^2}}}.\]
Substituting the first derivative \(y'\) into this formula, we have
\[y^{\prime\prime} = \frac{{xy' - y}}{{{y^2}}} = \frac{{x\left( { - \frac{x}{y}} \right) - y}}{{{y^2}}} = \frac{{ - \frac{{{x^2}}}{y} - y}}{{{y^2}}} = \frac{{ - {x^2} - {y^2}}}{{{y^3}}} = - \frac{{{x^2} + {y^2}}}{{{y^3}}} = - \frac{{{R^2}}}{{{y^3}}}.\]
Example 13.
The function \(y = f\left( x \right)\) is given in parametric form by the equations \[x = {t^3},\;y = {t^2} + 1,\] where \(t \gt 0.\) Find \(y_{xx}^{\prime\prime}.\)
Solution.
Determine the first derivative \(y_x^\prime:\)
\[y_x^\prime = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{\left( {{t^2} + 1} \right)_t^\prime}}{{\left( {{t^3}} \right)_t^\prime}} = \frac{{2t}}{{3{t^2}}} = \frac{2}{{3t}}.\]
Differentiate \(y_x^\prime\) again with respect to \(x:\)
\[y_{xx}^{\prime\prime} = \left( {y_x^\prime} \right)_x^\prime = \left( {y_x^\prime} \right)_t^\prime \cdot t_x^\prime = \left( {\frac{2}{{3t}}} \right)_t^\prime \cdot \frac{1}{{x_t^\prime}} = \frac{2}{3}\left( {{t^{ - 1}}} \right)_t^\prime \cdot \frac{1}{{x_t^\prime}} = \frac{2}{3} \cdot \left( { - 1} \right){t^{ - 2}} \cdot \frac{1}{{3{t^2}}} = - \frac{2}{{3{t^2}}} \cdot \frac{1}{{3{t^2}}} = - \frac{2}{{9{t^4}}}.\]
Example 14.
Find the second derivative of the function given by the equation \[{x^3} + {y^3} = 1.\]
Solution.
We use implicit differentiation:
\[{x^3} + {y^3} = 1,\;\; \Rightarrow \left( {{x^3}} \right)^\prime + \left( {{y^3}} \right)^\prime = 1^\prime,\;\; \Rightarrow 3{x^2} + 3{y^2}y^\prime = 0,\;\; \Rightarrow {x^2} + {y^2}y^\prime = 0,\,\; \Rightarrow y^\prime = - \frac{{{x^2}}}{{{y^2}}}.\]
Differentiate again the equation \({x^2} + {y^2}y^\prime = 0:\)
\[{x^2} + {y^2}y^\prime = 0,\;\; \Rightarrow \left( {{x^2}} \right)^\prime + \left( {{y^2}y'} \right)^\prime = 0,\;\; \Rightarrow 2x + 2yy^\prime y^\prime + {y^2}y^{\prime\prime} = 0,\;\; \Rightarrow 2x + 2y{\left( {y^\prime} \right)^2} + {y^2}y^{\prime\prime} = 0,\;\; \Rightarrow y^{\prime\prime} = - \frac{{2x + 2y{{\left( {y^\prime} \right)}^2}}}{{{y^2}}}.\]
Substitute the expression for the first derivative \(y^\prime\) found above:
\[y^{\prime\prime} = - \frac{{2x + 2y{{\left( {y^\prime} \right)}^2}}}{{{y^2}}} = - \frac{{2x + 2y{{\left( { - \frac{{{x^2}}}{{{y^2}}}} \right)}^2}}}{{{y^2}}} = - \frac{{2x + 2y \cdot \frac{{{x^4}}}{{{y^4}}}}}{{{y^2}}} = - \frac{{2x + \frac{{2{x^4}}}{{{y^3}}}}}{{{y^2}}} = - \frac{{\frac{{2x{y^3} + 2{x^4}}}{{{y^3}}}}}{{{y^2}}} = - \frac{{2{x^4} + 2x{y^3}}}{{{y^5}}} = - \frac{{2x\left( {{x^3} + {y^3}} \right)}}{{{y^5}}} = - \frac{{2x \cdot 1}}{{{y^5}}} = - \frac{{2x}}{{{y^5}}}.\]
Example 15.
The function \(y = f\left( x \right)\) is given in parametric form by the equations \[x = t + \cos t,\;y = 1 + \sin t,\] where \(t \in \left( {0,2\pi } \right).\) Find \(y_{xx}^{\prime\prime}.\)
Solution.
Taking the first derivative of the parametric function, we have
\[y_x^\prime = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{\left( {1 + \sin t} \right)_t^\prime}}{{\left( {t + \cos t} \right)_t^\prime}} = \frac{{\cos t}}{{1 - \sin t}}.\]
Now we differentiate both sides of the expression for \(y_x^\prime\) with respect to \(x.\) This yields:
\[y_{xx}^{\prime\prime} = \left( {y_x^\prime} \right)_x^\prime = \left( {y_x^\prime} \right)_t^\prime \cdot t_x^\prime = \left( {\frac{{\cos t}}{{1 - \sin t}}} \right)_t^\prime \cdot t_x^\prime = \left( {\frac{{\cos t}}{{1 - \sin t}}} \right)_t^\prime \cdot \frac{1}{{x_t^\prime}} = \frac{{\left( { - \sin t} \right) \left( {1 - \sin t} \right) - \cos t \left( { - \cos t} \right)}}{{{{\left( {1 - \sin t} \right)}^2}}} \cdot \frac{1}{{1 - \sin t}} = \frac{{ - \sin t + {{\sin }^2}t + {{\cos }^2}t}}{{{{\left( {1 - \sin t} \right)}^3}}} = \frac{{1 - \sin t}}{{{{\left( {1 - \sin t} \right)}^3}}} = \frac{1}{{{{\left( {1 - \sin t} \right)}^2}}}.\]
Example 16.
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:
\[y^{\prime\prime} = {\left( {\frac{{{e^{x - y}} - 1}}{{{e^{x - y}} + 1}}} \right)^\prime } = \frac{{2{e^{x - y}}\left( {1 - y'} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}}.\]
Substitute the expression for the first derivative:
\[y^{\prime\prime} = \frac{{2{e^{x - y}}\left( {1 - y'} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} = \frac{{2{e^{x - y}}\left( {1 - \frac{{{e^{x - y}} - 1}}{{{e^{x - y}} + 1}}} \right)}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} = \frac{{2{e^{x - y}} \cdot \frac{{\cancel{e^{x - y}} + 1 - \cancel{e^{x - y}} + 1}}{{{e^{x - y}} + 1}}}}{{{{\left( {{e^{x - y}} + 1} \right)}^2}}} = \frac{{4{e^{x - y}}}}{{{{\left( {{e^{x - y}} + 1} \right)}^3}}}.\]
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 \(y^{\prime\prime}:\)
\[y^{\prime\prime} = \frac{{4{e^{x - y}}}}{{{{\left( {{e^{x - y}} + 1} \right)}^3}}} = \frac{{4\left( {x + y} \right)}}{{{{\left( {x + y + 1} \right)}^3}}}.\]
