Higher-Order Derivatives
Higher-Order Derivatives of an Explicit Function
Let the function y = f (x) have a finite derivative f '(x) in a certain interval (a, b), i.e. the derivative f '(x) is also a function in this interval. If this function is differentiable, we can find the second derivative of the original function y = f (x), which is denoted by various notations as
\[f^{\prime\prime} = \left({f^\prime}\right)^\prime = {\left( {\frac{{dy}}{{dx}}} \right)^\prime } = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{{d^2}y}}{{d{x^2}}}.\]
Similarly, if f '' exists and is differentiable, we can calculate the third derivative of the function f (x):
\[f^{\prime\prime\prime} = \frac{{{d^3}y}}{{d{x^3}}} = y^{\prime\prime\prime}.\]
The result of taking the derivative \(n\) times is called the \(n\)th derivative of \(f\left( x \right)\) with respect to \(x\) and is denoted as
\[\frac{{{d^n}f}}{{d{x^n}}} = \frac{{{d^n}y}}{{d{x^n}}}\;\;\;\left( {\text{in Leibnitz's notation}} \right),\]
\[{f^{\left( n \right)}}\left( x \right) = {y^{\left( n \right)}}\left( x \right)\;\;\;\left( {\text{in Lagrange's notation}} \right).\]
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
\[y' = {f_1}\left( {x,y} \right),\;\;\;y^{\prime\prime} = {f_2}\left( {x,y} \right), \ldots,\;\;\;{y^{\left( n \right)}} = {f_n}\left( {x,y} \right).\]
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:
\[y^{\prime\prime} = {y^{\prime\prime}_{xx}} = \frac{{{\left( {{y'_x}} \right)}'_t}}{{{x'_t}}}.\]
Similarly, we define the derivatives of the third and higher order:
\[y^{\prime\prime\prime} = {y^{\prime\prime\prime}_{xxx}} = \frac{{{{\left( {{y^{\prime\prime}_{xx}}} \right)}'_t}}}{{{x'_t}}}, \ldots,\;\;y^{\left( n \right)} = y_{\underbrace {xx \ldots x}_n}^{\left( n \right)} = \frac{{{{\left( {y_{\underbrace {xx \ldots x}_{n - 1}}^{\left( {n - 1} \right)}} \right)}'_t}}}{{{x'_t}}}. \]
Solved Problems
Example 1.
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:
\[y = {\left( {2x + 1} \right)^3}\left( {x - 1} \right) = \left( {8{x^3} + 12{x^2} + 6x + 1} \right)\cdot \left( {x - 1} \right) = \color{blue}{8{x^4}} + \color{red}{12{x^3}} + \color{maroon}{6{x^2}} + \color{green}x - \color{red}{8{x^3}} - \color{maroon}{12{x^2}} - \color{green}{6x} - \color{coral}1 = \color{blue}{8{x^4}} + \color{red}{4{x^3}} - \color{maroon}{6{x^2}} - \color{green}{5x} - \color{coral}1.\]
Now we successively calculate the derivatives from \(1\)st to \(5\)th order:
\[y' = \left( {8{x^4} + 4{x^3} - 6{x^2} - 5x - 1} \right)^\prime = 32{x^3} + 12{x^2} - 12x - 5,\]
\[y^{\prime\prime} = {\left( {y'} \right)^\prime } = {\left( {32{x^3} + 12{x^2} - 12x - 5} \right)^\prime } = 96{x^2} + 24x - 12,\]
\[y^{\prime\prime\prime} = {\left( {y^{\prime\prime}} \right)^\prime } = {\left( {96{x^2} + 24x - 12} \right)^\prime } = 192x + 24,\]
\[{y^{\left( 4 \right)}} = {\left( {y^{\prime\prime\prime}} \right)^\prime } = {\left( {192x + 24} \right)^\prime } = 192,\]
\[{y^{\left( 5 \right)}} = {\left( {{y^{\left( 4 \right)}}} \right)^\prime } = {\left( {192} \right)^\prime } = 0.\]
Example 2.
Find the \(n\)th order derivative of the natural logarithm function \[y = \ln x.\]
Solution.
We calculate several successive derivatives of the given function:
\[y' = \left( {\ln x} \right)^\prime = \frac{1}{x},\]
\[y^{\prime\prime} = {\left( {y'} \right)^\prime } = {\left( {\frac{1}{x}} \right)^\prime } = {\left( {{x^{ - 1}}} \right)^\prime } = - {x^{ - 2}} = - \frac{1}{{{x^2}}},\]
\[y^{\prime\prime\prime} = {\left( {y^{\prime\prime}} \right)^\prime } = {\left( { - \frac{1}{{{x^2}}}} \right)^\prime } = 2{x^{ - 3}} = \frac{2}{{{x^3}}},\]
\[{y^{\left( 4 \right)}} = {\left( {y^{\prime\prime\prime}} \right)^\prime } = {\left( {\frac{2}{{{x^3}}}} \right)^\prime } = - 6{x^{ - 4}} = - \frac{6}{{{x^4}}},\]
\[{y^{\left( 5 \right)}} = {\left( {{y^{\left( 4 \right)}}} \right)^\prime } = {\left( { - \frac{6}{{{x^4}}}} \right)^\prime } = 24{x^{ - 5}} = \frac{{24}}{{{x^5}}}.\]
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 3.
Find the \(n\)th derivative of the function \[y = \ln \left( {ax + b} \right).\]
Solution.
Let's calculate the few first derivatives using the chain and power rules:
\[y^\prime = \left( {\ln \left( {ax + b} \right)} \right)^\prime = \frac{1}{{ax + b}} \cdot \left( {ax + b} \right)^\prime = \frac{a}{{ax + b}};\]
\[y^{\prime\prime} = \left( {\frac{a}{{ax + b}}} \right)^\prime = \left( {a{{\left( {ax + b} \right)}^{ - 1}}} \right)^\prime = - 1 \cdot {a^2}{\left( {ax + b} \right)^{ - 2}} = - \frac{{1 \cdot {a^2}}}{{{{\left( {ax + b} \right)}^2}}};\]
\[y^{\prime\prime\prime} = \left( { - 1 \cdot {a^2}{{\left( {ax + b} \right)}^{ - 2}}} \right)^\prime = - 1 \cdot \left( { - 2} \right) \cdot {a^3}{\left( {ax + b} \right)^{ - 3}} = 2!\,{a^3}{\left( {ax + b} \right)^{ - 3}} = \frac{{2!\,{a^3}}}{{{{\left( {ax + b} \right)}^3}}};\]
\[{y^{\left( 4 \right)}} = \left( {2!{a^3}{{\left( {ax + b} \right)}^{ - 3}}} \right)^\prime = 2! \cdot \left( { - 3} \right) \cdot {a^4}{\left( {ax + b} \right)^{ - 4}} = - 3!\,{a^4}{\left( {ax + b} \right)^{ - 4}} = - \frac{{3!\,{a^4}}}{{{{\left( {ax + b} \right)}^4}}}.\]
We can easily detect the common pattern, so the \(n\)th derivative is given by
\[y^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^{n - 1}}\left( {n - 1} \right)!\,{a^n}}}{{{{\left( {ax + b} \right)}^n}}}.\]
Example 4.
Find the \(n\)th derivative of the function \[y = x\ln x\] at \(x = 1.\)
Solution.
The first two derivatives are written as
\[y^\prime = \left( {x\ln x} \right)^\prime = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1;\]
\[y^{\prime\prime} = \left( {\ln x + 1} \right)^\prime = \frac{1}{x} = {x^{ - 1}}.\]
We can continue differentiating using the power rule:
\[y^{\prime\prime\prime} = \left( {{x^{ - 1}}} \right)^\prime = - 1 \cdot {x^{ - 2}} = - \frac{1}{{{x^2}}};\]
\[{y^{\left( 4 \right)}} = \left( { - 1 \cdot {x^{ - 2}}} \right)^\prime = 2!\,{x^{ - 3}} = \frac{{2!}}{{{x^3}}};\]
\[{y^{\left( 5 \right)}} = \left( {2!\,{x^{ - 3}}} \right)^\prime = 2! \cdot \left( { - 3} \right) \cdot {x^{ - 4}} = - 3!\,{x^{ - 4}} = - \frac{{3!}}{{{x^4}}}.\]
We see that the \(n\)th derivative \(\left(\text{at } {n \ge 2} \right)\) has the form
\[y^{\left( n \right)} = \frac{{{{\left( { - 1} \right)}^n}\left( {n - 1} \right)!}}{{{x^{n - 1}}}}.\]
Substituting \(x = 1\) yields:
\[{y^{\left( n \right)}}\left( 1 \right) = \frac{{{{\left( { - 1} \right)}^n}\left( {n - 1} \right)!}}{{{1^{n - 1}}}} = {\left( { - 1} \right)^n}\left( {n - 1} \right)!\]
Example 5.
Find all derivatives of the sine function.
Solution.
We calculate a few derivatives starting from the first one:
\[y' = {\left( {\sin x} \right)^\prime } = \cos x = \sin \left( {x + \frac{\pi }{2}} \right),\]
\[y^{\prime\prime} = {\left( {\cos x} \right)^\prime } = - \sin x = \sin \left( {x + 2 \cdot \frac{\pi }{2}} \right),\]
\[y^{\prime\prime\prime} = {\left( { - \sin x} \right)^\prime } = - \cos x = \sin \left( {x + 3 \cdot \frac{\pi }{2}} \right),\]
\[{y^{IV}} = {\left( { - \cos x} \right)^\prime } = \sin x = \sin \left( {x + 4 \cdot \frac{\pi }{2}} \right).\]
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 6.
Find the \(n\)th derivative of the function \[y = \sin ax.\]
Solution.
Differentiating successively and using the trig cofunction identities, we have
\[y^\prime = \left( {\sin ax} \right)^\prime = a\cos ax = a\sin \left( {ax + \frac{\pi }{2}} \right);\]
\[y^{\prime\prime} = \left( {a\cos ax} \right)^\prime = - {a^2}\sin ax = {a^2}\sin \left( {ax + 2 \cdot \frac{\pi }{2}} \right);\]
\[y^{\prime\prime\prime} = \left( { - {a^2}\sin ax} \right)^\prime = - {a^3}\cos ax = {a^3}\sin \left( {ax + 3 \cdot \frac{\pi }{2}} \right);\]
\[{y^{\left( 4 \right)}} = \left( { - {a^3}\cos ax} \right)^\prime = {a^4}\sin ax = {a^4}\sin \left( {ax + 2\pi } \right) = {a^4}\sin \left( {ax + 4 \cdot \frac{\pi }{2}} \right).\]
Then, the \(n\)th derivative is written in the form
\[y^{\left( n \right)} = {a^n}\sin \left( {ax + \frac{{\pi n}}{2}} \right).\]
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:
\[y' = {\left( {\cos x} \right)^\prime } = - \sin x = \cos \left( {x + \frac{\pi }{2}} \right),\]
\[y^{\prime\prime} = {\left( { - \sin x} \right)^\prime } = - \cos x = \cos \left( {x + 2 \cdot \frac{\pi }{2}} \right),\]
\[y^{\prime\prime\prime} = {\left( { - \cos x} \right)^\prime } = \sin x = \cos \left( {x + 3 \cdot \frac{\pi }{2}} \right),\]
\[{y^{IV}} = {\left( {\sin x} \right)^\prime } = \cos x = \cos \left( {x + 4 \cdot \frac{\pi }{2}} \right).\]
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 the \(n\)th order derivative of the function \[y = {\sin ^2}x.\]
Solution.
Let's calculate the first few derivatives:
\[y^\prime = \left( {{{\sin }^2}x} \right)^\prime = 2\sin x\cos x = \sin 2x;\]
\[y^{\prime\prime} = \left( {\sin 2x} \right)^\prime = 2\cos 2x = 2\sin \left( {2x + \frac{\pi }{2}} \right);\]
\[y^{\prime\prime\prime} = \left( {2\cos 2x} \right)^\prime = - 4\sin 2x = {2^2}\sin \left( {2x + \frac{\pi }{2} \cdot 2} \right);\]
\[{y^{\left( 4 \right)}} = \left( { - 4\sin 2x} \right)^\prime = - 8\cos 2x = {2^3}\sin \left( {2x + \frac{\pi }{2} \cdot 3} \right).\]
We used here cofunction identities:
\[\sin \left( {\alpha + \frac{\pi }{2}} \right) = \cos \alpha ;\]
\[\sin \left( {\alpha + \pi } \right) = - \sin \alpha ;\]
\[\sin \left( {\alpha + \frac{{3\pi }}{2}} \right) = - \cos \alpha .\]
It is easy to detect the following pattern:
\[{y^{\left( n \right)}} = {2^{n - 1}}\sin \left( {2x + \frac{{\pi n}}{2}} \right),\]
where \(n = 1,2,3, \ldots \)
See more problems on Page 2.
