Higher-Order Derivatives
Solved Problems
Example 9.
Find all derivatives of the function \[y = {\frac{1}{x}}.\]
Solution.
We find a few first derivatives:
\[y' = {\left( {\frac{1}{x}} \right)^\prime } = - \frac{1}{{{x^2}}},\]
\[y^{\prime\prime} = \left( { - \frac{1}{{{x^2}}}} \right)^\prime = - 1 \cdot {\left( {{x^{ - 2}}} \right)^\prime } = - 1 \cdot \left( { - 2} \right) \cdot {x^{ - 3}} = \frac{{{{\left( { - 1} \right)}^2} \cdot 2}}{{{x^3}}},\]
\[y^{\prime\prime\prime} = \left( {\frac{{{{\left( { - 1} \right)}^2} \cdot 2}}{{{x^3}}}} \right)^\prime = {\left( { - 1} \right)^2} \cdot 2 \cdot {\left( {{x^{ - 3}}} \right)^\prime } = {\left( { - 1} \right)^3} \cdot 2 \cdot 3 \cdot {x^{ - 4}} = \frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 3}}{{{x^4}}},\]
\[{y^{IV}} = \left( {\frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 3}}{{{x^4}}}} \right)^\prime = {\left( { - 1} \right)^3} \cdot 2 \cdot 3 \cdot {\left( {{x^{ - 4}}} \right)^\prime } = {\left( { - 1} \right)^4} \cdot 4! \cdot {x^{ - 5}} = \frac{{{{\left( { - 1} \right)}^4}4!}}{{{x^5}}}.\]
This is enough to detect the general pattern:
\[{y^{\left( n \right)}} = \frac{{{{\left( { - 1} \right)}^n}n!}}{{{x^{n + 1}}}}.\]
Example 10.
Find all derivatives of the function \[y = \frac{1}{{2x - 3}}.\]
Solution.
We take the few first derivatives:
\[y^\prime = \left( {\frac{1}{{2x - 3}}} \right)^\prime = \left( {{{\left( {2x - 3} \right)}^{ - 1}}} \right)^\prime = - {\left( {2x - 3} \right)^{ - 2}} \cdot \left( {2x - 3} \right)^\prime = - 1 \cdot {\left( {2x - 3} \right)^{ - 2}} \cdot 2 = - \frac{2}{{{{\left( {2x - 3} \right)}^2}}};\]
\[y^{\prime\prime} = \left( { - 1 \cdot {{\left( {2x - 3} \right)}^{ - 2}} \cdot 2} \right)^\prime = - 1 \cdot \left( { - 2} \right) \cdot {\left( {2x - 3} \right)^{ - 3}} \cdot {2^2} = 2!\,{\left( {2x - 3} \right)^{ - 3}}{2^2} = \frac{{2!\,{2^2}}}{{{{\left( {2x - 3} \right)}^3}}};\]
\[y^{\prime\prime\prime} = \left( {2!\,{{\left( {2x - 3} \right)}^{ - 3}}{2^2}} \right)^\prime = 2! \cdot \left( { - 3} \right) \cdot {\left( {2x - 3} \right)^{ - 4}} \cdot {2^3} = - 3!\,{\left( {2x - 3} \right)^{ - 4}}{2^3} = - \frac{{3!\,{2^3}}}{{{{\left( {2x - 3} \right)}^4}}};\]
\[{y^{\left( 4 \right)}} = \left( { - 3!\,{{\left( {2x - 3} \right)}^{ - 4}}{2^3}} \right)^\prime = - 3! \cdot \left( { - 4} \right) \cdot {\left( {2x - 3} \right)^{ - 5}} \cdot {2^4} = 4!\,{\left( {2x - 3} \right)^{ - 5}}{2^4} = \frac{{4!\,{2^4}}}{{{{\left( {2x - 3} \right)}^5}}}.\]
It is clear that the \(n\)th order derivative is written as
\[y^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^n}n!\,{2^n}}}{{{{\left( {2x - 3} \right)}^{n + 1}}}}.\]
Example 11.
Find all derivatives of the function \[y = \frac{1}{{1 - 5x}}.\]
Solution.
Let's take the few first derivatives using the power rule and the chain rule:
\[y^\prime = \left( {\frac{1}{{1 - 5x}}} \right)^\prime = \left( {{{\left( {1 - 5x} \right)}^{ - 1}}} \right)^\prime = - 1 \cdot {\left( {1 - 5x} \right)^{ - 2}} \cdot \left( {1 - 5x} \right)^\prime = - 1 \cdot {\left( {1 - 5x} \right)^{ - 2}} \cdot \left( { - 5} \right) = 1 \cdot {\left( {1 - 5x} \right)^{ - 2}} \cdot 5 = \frac{{1 \cdot 5}}{{{{\left( {1 - 5x} \right)}^2}}};\]
\[y^{\prime\prime} = \left( {1 \cdot {{\left( {1 - 5x} \right)}^{ - 2}} \cdot 5} \right)^\prime = 1 \cdot \left( { - 2} \right) \cdot {\left( {1 - 5x} \right)^{ - 3}} \cdot 5 \cdot \left( { - 5} \right) = 2! \cdot {\left( {1 - 5x} \right)^{ - 3}} \cdot {5^2} = \frac{{2!\,{5^2}}}{{{{\left( {1 - 5x} \right)}^3}}};\]
\[y^{\prime\prime\prime} = \left( {2! \cdot {{\left( {1 - 5x} \right)}^{ - 3}} \cdot {5^2}} \right)^\prime = 2! \cdot \left( { - 3} \right) \cdot {\left( {1 - 5x} \right)^{ - 4}} \cdot {5^2} \cdot \left( { - 5} \right) = 3! \cdot {\left( {1 - 5x} \right)^{ - 4}} \cdot {5^3} = \frac{{3!\,{5^3}}}{{{{\left( {1 - 5x} \right)}^4}}};\]
\[{y^{\left( 4 \right)}} = \left( {3! \cdot {{\left( {1 - 5x} \right)}^{ - 4}} \cdot {5^3}} \right)^\prime = 3! \cdot \left( { - 4} \right) \cdot {\left( {1 - 5x} \right)^{ - 5}} \cdot {5^3} \cdot \left( { - 5} \right) = 4! \cdot {\left( {1 - 5x} \right)^{ - 5}} \cdot {5^4} = \frac{{4!\,{5^4}}}{{{{\left( {1 - 5x} \right)}^5}}}.\]
So, the derivative of the \(n\)th order is given by
\[{y^{\left( n \right)}} = \frac{{n!\,{5^n}}}{{{{\left( {1 - 5x} \right)}^{n + 1}}}}.\]
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.
\[y' = {\left( {{3^{2x + 1}}} \right)^\prime } = {3^{2x + 1}} \cdot \ln 3 \cdot {\left( {2x + 1} \right)^\prime } = {3^{2x + 1}} \cdot 2\ln 3,\]
\[y^{\prime\prime} = {\left( {y'} \right)^\prime } = {\left( {{3^{2x + 1}} \cdot 2\ln 3} \right)^\prime } = {\left( {{3^{2x + 1}}} \right)^\prime } \cdot 2\ln 3 = {3^{2x + 1}} \cdot {2^2}{\ln ^2}3,\]
\[y^{\prime\prime\prime} = {\left( {y^{\prime\prime}} \right)^\prime } = {\left( {{3^{2x + 1}} \cdot {2^2}{{\ln }^2}3} \right)^\prime } = {\left( {{3^{2x + 1}}} \right)^\prime } \cdot {2^2}{\ln ^2}3 = {3^{2x + 1}} \cdot {2^3}{\ln ^3}3.\]
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 for \(n = k:\)
\[{y^{\left( k \right)}} = {\left( {{3^{2x + 1}}} \right)^{\left( k \right)}} = {3^{2x + 1}} \cdot {2^k}\,{\ln ^k}3.\]
Then for \(n = k + 1,\) we have
\[{y^{\left( {k + 1} \right)}} = {\left( {{3^{2x + 1}}} \right)^{\left( {k + 1} \right)}} = {\left[ {{{\left( {{3^{2x + 1}}} \right)}^{\left( k \right)}}} \right]^\prime } = {\left( {{3^{2x + 1}} \cdot {2^k}\,{{\ln }^k}3} \right)^\prime } = {\left( {{3^{2x + 1}}} \right)^\prime } \cdot {2^k}\,{\ln ^k}3 = {3^{2x + 1}} \cdot {2^{k + 1}}{\ln ^{k + 1}}3,\]
that is, this formula holds for \(n = k + 1.\) Consequently, it is true for any natural number \(n.\)
Example 13.
Find the \(n\)th derivative of the function \[y = x{e^x}\] at \(x = 0.\)
Solution.
We differentiate successively using the product rule:
\[y^\prime = \left( {x{e^x}} \right)^\prime = 1 \cdot {e^x} + x \cdot {e^x} = \left( {x + 1} \right){e^x};\]
\[y^{\prime\prime} = \left( {\left( {x + 1} \right){e^x}} \right)^\prime = 1 \cdot {e^x} + \left( {x + 1} \right) \cdot {e^x} = \left( {x + 2} \right){e^x};\]
\[y^{\prime\prime\prime} = \left( {\left( {x + 2} \right){e^x}} \right)^\prime = 1 \cdot {e^x} + \left( {x + 2} \right) \cdot {e^x} = \left( {x + 3} \right){e^x}.\]
Hence
\[y^{\left( n \right)} = \left( {x + n} \right){e^x},\]
where \(n\) is a whole number. Then
\[{y^{\left( n \right)}}\left( 0 \right) = \left( {0 + n} \right){e^0} = n \cdot 1 = n.\]
Example 14.
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:
\[y' = {\left( {{x^m}} \right)^\prime } = m{x^{m - 1}},\]
\[y^{\prime\prime} = \left( {y'} \right)^\prime = \left( {m{x^{m - 1}}} \right)^\prime = m\left( {m - 1} \right){x^{m - 2}},\]
\[y^{\prime\prime\prime} = \left( {y^{\prime\prime}} \right)^\prime = \left[ {m\left( {m - 1} \right){x^{m - 2}}} \right]^\prime = \left[ {m\left( {m - 1} \right)\left( {m - 2} \right)} \right]{x^{m - 3}}.\]
Hence, it is easy to establish a general expression for the \(n\)th order derivative:
\[y^{\left( n \right)} = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n + 1} \right){x^{m - n}}.\]
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:
\[y^{\left( {n + 1} \right)} = {\left[ {{y^{\left( n \right)}}} \right]^\prime } = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n + 1} \right){\left( {{x^{m - n}}} \right)^\prime } = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n} \right){x^{m - \left( {n + 1} \right)}}.\]
As you can see, 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 \(n \gt m\) are equal to zero.
Example 15.
Find the \(n\)th order derivative of the square root \[y = \sqrt x .\]
Solution.
We use the result of Example \(14\), where the \(n\)th order derivative of the power function with an arbitrary real exponent \(m\) is derived. In this case we have
\[y = \sqrt x = {x^{\frac{1}{2}}}\;\;\left( {m = \frac{1}{2}} \right).\]
Then the \(n\)th order derivative is written as
\[y^{\left( n \right)} = \left( {{x^{\frac{1}{2}}}} \right)^{\left( n \right)} = \frac{1}{2} \cdot \left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right) \cdots \left( {\frac{1}{2} - n + 1} \right){x^{\frac{1}{2} - n}} = \frac{1}{2} \cdot \left( { - \frac{1}{2}} \right) \cdot \left( { - \frac{3}{2}} \right) \cdot \left( { - \frac{5}{2}} \right) \cdots \left[ { - \left( {n - \frac{3}{2}} \right)} \right]\frac{{{x^{\frac{1}{2}}}}}{{{x^n}}} = \frac{1}{2} \cdot {\left( { - 1} \right)^{n - 1}} \left[ {\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2} \cdots \frac{{2n - 3}}{2}} \right]\frac{{\sqrt x }}{{{x^n}}} = \frac{1}{2} \cdot {\left( { - 1} \right)^{n - 1}} \cdot \frac{1}{{{2^{n - 1}}}} \cdot \left[ {1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right)} \right]\frac{{\sqrt x }}{{{x^n}}} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left[ {1 \cdot 3 \cdot 5 \cdots \left( {2n - 3} \right)} \right]\sqrt x }}{{{x^n}}}.\]
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,
\[y^{\prime\prime} = \frac{{{{\left( { - 1} \right)}^1}1!!\sqrt x }}{{{{\left( {2x} \right)}^2}}} = - \frac{{\sqrt x }}{{4{x^2}}} = - \frac{1}{{4\sqrt {{x^3}} }},\]
\[y^{\prime\prime\prime} = \frac{{{{\left( { - 1} \right)}^2}3!!\sqrt x }}{{{{\left( {2x} \right)}^3}}} = \frac{{3\sqrt x }}{{8{x^3}}} = \frac{3}{{8\sqrt {{x^5}} }}.\]
Example 16.
Find the \(n\)th order derivative of the cube root \[y = \sqrt[3]{x}.\]
Solution.
The first derivative of the cube root is given by
\[y' = \left( {\sqrt[3]{x}} \right)^\prime = \left( {{x^{\frac{1}{3}}}} \right)^\prime = \frac{1}{3}{x^{\frac{1}{3} - 1}} = \frac{1}{3}{x^{ - \frac{2}{3}}} = \frac{1}{{3\sqrt[3]{{{x^2}}}}}.\]
Next, we use the general formula for the \(n\)th order derivative of the power function \(y = {x^m}\) (Example \(14\)):
\[\left( {{x^m}} \right)^{\left( n \right)} = m\left( {m - 1} \right)\left( {m - 2} \right) \cdots \left( {m - n + 1} \right){x^{m - n}}.\]
In our case \(m = {\frac{1}{3}}.\) Consequently, the derivative is as follows:
\[y^{\left( n \right)} = \left( {\sqrt[3]{x}} \right)^{\left( n \right)} = \left( {{x^{\frac{1}{3}}}} \right)^{\left( n \right)} = \frac{1}{3}\left( {\frac{1}{3} - 1} \right) \left( {\frac{1}{3} - 2} \right)\left( {\frac{1}{3} - 3} \right) \cdots \left( {\frac{1}{3} - n + 1} \right){x^{\frac{1}{3} - n}} = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right) \cdot \left( { - \frac{5}{3}} \right) \cdot \left( { - \frac{8}{3}} \right) \cdots \left[ { - \left( {n - \frac{4}{3}} \right)} \right]\frac{{{x^{\frac{1}{3}}}}}{{{x^n}}} = \frac{1}{3} \cdot {\left( { - 1} \right)^{n - 1}} \left[ {\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdots \frac{{3n - 4}}{3}} \right]\frac{{\sqrt[3]{x}}}{{{x^n}}} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left[ {2 \cdot 5 \cdot 8 \cdots \left( {3n - 4} \right)} \right]\sqrt[3]{x}}}{{{{\left( {3x} \right)}^n}}},\]
where \(n \ge 2.\)
Specifically, the second and third derivatives of the cube root are expressed by the formulas
\[y^{\prime\prime} = \frac{{{{\left( { - 1} \right)}^1} \cdot 2 \cdot \sqrt[3]{x}}}{{{{\left( {3x} \right)}^2}}} = - \frac{{2\sqrt[3]{x}}}{{9{x^2}}} = - \frac{2}{{9\sqrt[3]{{{x^5}}}}},\]
\[y^{\prime\prime\prime} = \frac{{{{\left( { - 1} \right)}^2} \cdot 2 \cdot 5 \cdot \sqrt[3]{x}}}{{{{\left( {3x} \right)}^3}}} = \frac{{10\sqrt[3]{x}}}{{27{x^3}}} = \frac{{10}}{{27\sqrt[3]{{{x^8}}}}}.\]
Example 17.
Find the \(3\)rd derivative of the function \(y = f\left( x \right)\) given by the equation \[{y^2} = 2x.\]
Solution.
Differentiate both sides with respect to \(x\) to calculate the first derivative \(y_x^\prime:\)
\[\left( {{y^2}} \right)^\prime = \left( {2x} \right)^\prime,\;\; \Rightarrow 2yy^\prime = 2,\;\; \Rightarrow yy^\prime = 1,\;\; \Rightarrow y^\prime = y_x^\prime = \frac{1}{y}.\]
The second derivative \(y_{xx}^{\prime\prime}\) is given by
\[\left( {yy^\prime} \right)^\prime = 1^\prime,\;\; \Rightarrow y^\prime y^\prime + yy^{\prime\prime} = 0,\;\; \Rightarrow yy^{\prime\prime} = - \left( {y^\prime} \right)^2,\;\; \Rightarrow y^{\prime\prime} = y_{xx}^{\prime\prime} = - \frac{{{{\left( {y^\prime} \right)}^2}}}{y} = - \frac{{{{\left( {\frac{1}{y}} \right)}^2}}}{y} = - \frac{1}{{{y^3}}}.\]
Differentiating once more with respect to \(x,\) we find the \(3\)rd derivative \(y_{xxx}^{\prime\prime\prime}:\)
\[\left( {{{\left( {y^\prime} \right)}^2} + yy^{\prime\prime}} \right)^\prime = 0^\prime,\;\; \Rightarrow 2y^\prime y^{\prime\prime} + y^\prime y^{\prime\prime} + yy^{\prime\prime\prime} = 0,\;\; \Rightarrow 3y^\prime y^{\prime\prime} + yy^{\prime\prime\prime} = 0,\;\; \Rightarrow y^{\prime\prime\prime} = y_{xxx}^{\prime\prime\prime} = - \frac{{3y^\prime y^{\prime\prime}}}{y} = - \frac{{3 \cdot \frac{1}{y} \cdot \left( { - \frac{1}{{{y^3}}}} \right)}}{y} = \frac{3}{{{y^5}}}.\]
Example 18.
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:
\[y^\prime = {y'_x} = \frac{{{y'_t}}}{{{x'_t}}} = \frac{{{{\left( {b\sin t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} = \frac{{b\cos t}}{{ - a\sin t}} = - \frac{b}{a}\cot t,\]
\[y^{\prime\prime} = y^{\prime\prime}_{xx} = \frac{{{{\left( {{y'_x}} \right)}'_t}}}{{{x'_t}}} = \frac{{{{\left( { - \frac{b}{a}\cot t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} = \frac{{\left( { - \frac{b}{a}} \right)\left( { - \frac{1}{{{{\sin }^2}t}}} \right)}}{{\left( { - a\sin t} \right)}} = - \frac{b}{{{a^2}}}\frac{1}{{{{\sin }^3}t}} = - \frac{b}{{{a^2}}}{\csc ^3}t,\]
\[y^{\prime\prime\prime} = y^{\prime\prime\prime}_{xxx} = \frac{{{{\left( {{y^{\prime\prime}_{xx}}} \right)}'_t} }}{{{x'_t}}} = \frac{{{{\left( { - \frac{b}{{{a^2}}}{{\csc }^3}t} \right)}^\prime }}}{{{{\left( {a\cos t} \right)}^\prime }}} = \frac{{\left( { - \frac{b}{{{a^2}}}} \right) \cdot 3{{\csc }^2}t \cdot {{\left( {\csc t} \right)}^\prime }}}{{\left( { - a\sin t} \right)}} = - \frac{{3b}}{{{a^3}}} \cdot \frac{{{{\csc }^2}t \cdot \left( { - \cot t} \right) \cdot \csc t}}{{\left( { - \sin t} \right)}} = - \frac{{3b}}{{{a^3}}} \cdot \frac{{{{\csc }^3}t\cot t}}{{\sin t}} = - \frac{{3b}}{{{a^3}}}{\csc ^4}t\cot t.\]
Example 19.
Find the \(3\)rd derivative of the function given by the parametric equations \[x = 1 + {t^2},\; y = t - {t^3}\] at \(t = 1.\)
Take the first derivative:
\[y^\prime = y_x^\prime = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{\left( {t - {t^3}} \right)^\prime}}{{\left( {1 + {t^2}} \right)^\prime}} = \frac{{1 - 3{t^2}}}{{2t}} = \frac{1}{{2t}} - \frac{{3t}}{2}.\]
Continue differentiating:
\[y^{\prime\prime} = y_{xx}^{\prime\prime} = \frac{{\left( {y_x^\prime} \right)_t^\prime}}{{x_t^\prime}} = \frac{{\left( {\frac{1}{{2t}} - \frac{{3t}}{2}} \right)^\prime}}{{\left( {1 + {t^2}} \right)^\prime}} = \frac{{\frac{1}{2} \cdot \left( { - {t^{ - 2}}} \right) - \frac{3}{2}}}{{2t}} = - \frac{{\frac{1}{{2{t^2}}} + \frac{3}{2}}}{{2t}} = - \frac{1}{{4{t^3}}} - \frac{3}{{4t}}.\]
Similarly we calculate the third derivative \(y_{xxx}^{\prime\prime\prime}:\)
\[y^{\prime\prime\prime} = y_{xxx}^{\prime\prime\prime} = \frac{{\left( {y_{xx}^{\prime\prime}} \right)_t^\prime}}{{x_t^\prime}} = \frac{{\left( { - \frac{1}{{4{t^3}}} - \frac{3}{{4t}}} \right)^\prime}}{{\left( {1 + {t^2}} \right)^\prime}} = \frac{{ - \frac{1}{4} \cdot \left( { - 3{t^{ - 4}}} \right) - \frac{3}{4} \cdot \left( { - {t^{ - 2}}} \right)}}{{2t}} = \frac{{\frac{3}{{4{t^4}}} + \frac{3}{{4{t^2}}}}}{{2t}} = \frac{3}{{8{t^5}}} + \frac{3}{{8{t^3}}}.\]
At the point where \(t = 1,\) the third derivative is equal to
\[y_{xxx}^{\prime\prime\prime}\left( {t = 1} \right) = \frac{3}{{8 \cdot {1^5}}} + \frac{3}{{8 \cdot {1^3}}} = \frac{3}{8} + \frac{3}{8} = \frac{3}{4}.\]
Example 20.
Find the \(3\)rd derivative of the function given by the parametric equations \[x = 1 + \sin t,\;y = t - \cos t\] at \(t = 0.\)
Solution.
Compute the first derivative \(y_x^\prime:\)
\[y^\prime = y_x^\prime = \frac{{y_t^\prime}}{{x_t^\prime}} = \frac{{\left( {t - \cos t} \right)^\prime}}{{\left( {1 + \sin t} \right)^\prime}} = \frac{{1 + \sin t}}{{\cos t}}.\]
Similarly we determine the second derivative \(y_{xx}^{\prime\prime}\) of the parametric function:
\[y^{\prime\prime} = y_{xx}^{\prime\prime} = \frac{{\left( {y_x^\prime} \right)_t^\prime}}{{x_t^\prime}} = \frac{{\left( {\frac{{1 + \sin t}}{{\cos t}}} \right)^\prime}}{{\left( {1 + \sin t} \right)^\prime}} = \frac{{\frac{{\cos t \cdot \cos t - \left( {1 + \sin t} \right) \cdot \left( { - \sin t} \right)}}{{{{\cos }^2}t}}}}{{\cos t}} = \frac{{{{\cos }^2}t + \sin t + {{\sin }^2}t}}{{{{\cos }^3}t}} = \frac{{1 + \sin t}}{{{{\cos }^3}t}}.\]
The third derivative \(y_{xxx}^{\prime\prime\prime}\) has the form
\[y^{\prime\prime\prime} = y_{xxx}^{\prime\prime\prime} = \frac{{\left( {y_{xx}^{\prime\prime}} \right)_t^\prime}}{{x_t^\prime}} = \frac{{\left( {\frac{{1 + \sin t}}{{{{\cos }^3}t}}} \right)^\prime}}{{\left( {1 + \sin t} \right)^\prime}} = \frac{{{{\cos }^4}t + 3\,{{\cos }^2}t\sin t + 3\,{{\cos }^2}t\,{{\sin }^2}t}}{{{{\cos }^7}t}} = \frac{{{{\cos }^2}t\left( {{{\cos }^2}t + 3\sin t + 3\,{{\sin }^2}t} \right)}}{{{{\cos }^7}t}} = \frac{{{{\cos }^2}t + 3\sin t + 3\,{{\sin }^2}t}}{{{{\cos }^5}t}}.\]
So, at the point where \(t = 0,\) the third derivative is equal to
\[y_{xxx}^{\prime\prime\prime}\left( {t = 0} \right) = \frac{{{{\cos }^2}0 + 3\sin 0 + 3\,{{\sin }^2}0}}{{{{\cos }^5}0}} = \frac{{1 + 0 + 0}}{1} = 1.\]
Example 21.
Find the third derivative of the function given by the equation \[{x^2} + 3xy + {y^2} = 1.\]
Solution.
Differentiate one time with respect to \(x:\)
\[\left( {{x^2} + 3xy + {y^2}} \right)^\prime = {1^\prime },\;\; \Rightarrow 2x + 3\left( {x'y + xy'} \right) + 2yy' = 0,\;\; \Rightarrow 2x + 3y + 3xy' + 2yy' = 0,\;\; \Rightarrow 2x + 3y + \left( {3x + 2y} \right)y' = 0,\; \Rightarrow y' = - \frac{{2x + 3y}}{{3x + 2y}}.\]
Now we differentiate the last expression again considering \(y\) as a composite function:
\[y^{\prime\prime} = {\left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)^\prime } = \frac{{5y - 5xy'}}{{{{\left( {3x + 2y} \right)}^2}}}.\]
Substitute the explicit expression for the first derivative \(y':\)
\[y^{\prime\prime} = \frac{{5y - 5xy'}}{{{{\left( {3x + 2y} \right)}^2}}} = \frac{{5y - 5x\left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)}}{{{{\left( {3x + 2y} \right)}^2}}} = \frac{{5y\left( {3x + 2y} \right) + 5x\left( {2x + 3y} \right)}}{{{{\left( {3x + 2y} \right)}^3}}} = \frac{{15xy + 10{y^2} + 10{x^2} + 15xy}}{{{{\left( {3x + 2y} \right)}^3}}} = \frac{{10\left( {{x^2} + 3xy + {y^2}} \right)}}{{{{\left( {3x + 2y} \right)}^3}}}.\]
Since \({x^2} + 3xy + {y^2} = 1,\) we obtain the following expression for the second derivative \(y^{\prime\prime}:\)
\[y^{\prime\prime} = \frac{{10}}{{{{\left( {3x + 2y} \right)}^3}}}.\]
Similarly, differentiating once more, we get the third derivative:
\[y^{\prime\prime\prime} = {\left[ {\frac{{10}}{{{{\left( {3x + 2y} \right)}^3}}}} \right]^\prime } = - \frac{{30\left( {3 + 2y'} \right)}}{{{{\left( {3x + 2y} \right)}^4}}} = - \frac{{90 + 60y'}}{{{{\left( {3x + 2y} \right)}^4}}}.\]
We substitute again the first derivative and find:
\[y^{\prime\prime\prime} = - \frac{{90 + 60y'}}{{{{\left( {3x + 2y} \right)}^4}}} = - \frac{{90 + 60 \cdot \left( { - \frac{{2x + 3y}}{{3x + 2y}}} \right)}}{{{{\left( {3x + 2y} \right)}^4}}} = - \frac{{90\left( {3x + 2y} \right) - 60\left( {2x + 3y} \right)}}{{{{\left( {3x + 2y} \right)}^5}}} = - \frac{{150}}{{{{\left( {3x + 2y} \right)}^5}}}.\]
Example 22.
Find the \(3\)rd derivative of the function \(y = f\left( x \right)\) given by the equation \[{x^2} - {y^2} = 9.\]
Solution.
We differentiate both sides of the equation with respect to \(x\) keeping in mind that \(y\) is a function of \(x:\)
\[\left( {{x^2} - {y^2}} \right)^\prime = 9^\prime,\;\; \Rightarrow 2x - 2yy^\prime = 0, \Rightarrow x - yy^\prime = 0,\; \Rightarrow y^\prime = y_x^\prime = \frac{x}{y}.\]
Continue differentiating to obtain \(y_{xx}^{\prime\prime}:\)
\[x - yy^\prime = 0,\;\; \Rightarrow \left( {x - yy^\prime} \right)^\prime = 0,\;\; \Rightarrow 1 - y^\prime y^\prime - yy^{\prime\prime} = 0,\;\; \Rightarrow yy^{\prime\prime} = 1 - {\left( {y^\prime} \right)^2},\;\; \Rightarrow y^{\prime\prime} = y_{xx}^{\prime\prime} = \frac{{1 - {{\left( {y^\prime} \right)}^2}}}{y} = \frac{{1 - {{\left( {\frac{x}{y}} \right)}^2}}}{y} = \frac{{1 - \frac{{{x^2}}}{{{y^2}}}}}{y} = \frac{{{y^2} - {x^2}}}{{{y^3}}} = - \frac{{{x^2} - {y^2}}}{{{y^3}}} = - \frac{9}{{{y^3}}}.\]
Similarly we find the third derivative:
\[yy^{\prime\prime} = 1 - {\left( {y^\prime} \right)^2},\;\; \Rightarrow \left( {yy^{\prime\prime}} \right)^\prime = \left( {1 - {{\left( {y^\prime} \right)}^2}} \right)^\prime,\;\; \Rightarrow y^\prime y^{\prime\prime} + yy^{\prime\prime\prime} = - 2y^\prime y^{\prime\prime},\;\; \Rightarrow yy^{\prime\prime\prime} = - 3y^\prime y^{\prime\prime},\;\; \Rightarrow y^{\prime\prime\prime} = y_{xxx}^{\prime\prime\prime} = - \frac{{3y^\prime y^{\prime\prime}}}{y} = - \frac{{3 \cdot \frac{x}{y} \cdot \left( { - \frac{9}{{{y^3}}}} \right)}}{y} = \frac{{\frac{{27x}}{{{y^4}}}}}{y} = \frac{{27x}}{{{y^5}}}.\]
