# 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 $$y = f\left( x \right)$$, 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 $$f^{\prime\prime}$$ exists and is differentiable, we can calculate the third derivative of the function $$f\left( x \right):$$

${f^{\prime\prime\prime} = \frac{{{d^3}y}}{{d{x^3}}} }={ y^{\prime\prime\prime}.}$

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

${\frac{{{d^n}f}}{{d{x^n}}} }={ \frac{{{d^n}y}}{{d{x^n}}}\;\;\;}\kern0pt{\left( {\text{in Leibnitz’s notation}} \right),}$

${{f^{\left( n \right)}}\left( x \right) }={ {y^{\left( n \right)}}\left( x \right)\;\;\;}\kern0pt{\left( {\text{in Lagrange’s notation}} \right).}$

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:

${{\left( {u + v} \right)^{\left( n \right)}} = {u^{\left( n \right)}} + {v^{\left( n \right)}},\;\;\;}\kern-0.3pt {{\left( {Cu} \right)^{\left( n \right)}} = C{u^{\left( n \right)}},\;\;}\kern-0.3pt{C = \text{const}.}$

### 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

${y’ = {f_1}\left( {x,y} \right),\;\;\;}\kern-0.3pt {y^{\prime\prime} = {f_2}\left( {x,y} \right), \ldots,\;\;\;}\kern-0.3pt {{y^{\left( n \right)}} = {f_n}\left( {x,y} \right).}$

### 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:

${{y^{\prime\prime\prime} = {y^{\prime\prime\prime}_{xxx}} = \frac{{{{\left( {{y^{\prime\prime}_{xx}}} \right)}’_t}}}{{{x’_t}}}, \ldots,}\;\;}\kern0pt {{{y^{\left( n \right)}} = y_{\underbrace {xx \ldots x}_n}^{\left( n \right)} }={ \frac{{{{\left( {y_{\underbrace {xx \ldots x}_{n – 1}}^{\left( {n – 1} \right)}} \right)}’_t}}}{{{x’_t}}}.}}$

## Solved Problems

Click or tap a problem to see the solution.

### 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 $$n = 5.$$

### Example 2

Find the $$n$$th order derivative of the natural logarithm function $$y = \ln x.$$

### Example 3

Find the $$n$$th derivative of the function $$y = \ln \left( {ax + b} \right).$$

### Example 4

Find the $$n$$th derivative of the function $$y = x\ln x$$ at $$x = 1.$$

### Example 5

Find all derivatives of the sine function.

### Example 6

Find the $$n$$th derivative of the function $$y = \sin ax.$$

### Example 7

Find all derivatives of the cosine function.

### Example 8

Find the $$n$$th order derivative of the function $$y = {\sin ^2}x.$$

### Example 9

Find all derivatives of the function $$y = {\large\frac{1}{x}\normalsize}.$$

### Example 10

Find all derivatives of the function $$y = \large{\frac{1}{{2x – 3}}}\normalsize.$$

### Example 11

Find all derivatives of the function $$y = \large{\frac{1}{{1 – 5x}}}\normalsize.$$

### Example 12

Find the $$n$$th order derivative of the function $$y = {3^{2x + 1}}.$$

### Example 13

Find the $$n$$th derivative of the function $$y = x{e^x}$$ at $$x = 0.$$

### Example 14

Find the $$n$$th order derivative of the power function $$y = {x^m}$$ where $$m$$ is a real number.

### Example 15

Find the $$n$$th order derivative of the square root $$y = \sqrt x .$$

### Example 16

Find the $$n$$th order derivative of the cube root $$y = \sqrt[\large 3\normalsize]{x}.$$

### Example 17

Find the $$3$$rd derivative of the function $$y = f\left( x \right)$$ given by the equation $${y^2} = 2x.$$

### Example 18

Given the equation of an ellipse in parametric form: ${x = a\cos t,\;\;\;}\kern-0.3pt{y = b\sin t,}$ 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 $$x.$$

### Example 19

Find the $$3$$rd derivative of the function given by the parametric equations $$x = 1 + {t^2},$$ $$y = t – {t^3}$$ at $$t = 1.$$

### Example 20

Find the $$3$$rd derivative of the function given by the parametric equations $$x = 1 + \sin t,$$ $$y = t – \cos t$$ at $$t = 0.$$

### Example 21

Find the third derivative of the function given by the equation $${x^2} + 3xy + {y^2} = 1.$$

### Example 22

Find the $$3$$rd derivative of the function $$y = f\left( x \right)$$ given by the equation $${x^2} – {y^2} = 9.$$

### 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:

${y = {\left( {2x + 1} \right)^3}\left( {x – 1} \right) } = {\left( {8{x^3} + 12{x^2} + 6x + 1} \right)\cdot}\kern0pt{\left( {x – 1} \right) } = {\color{blue}{8{x^4}} + \color{red}{12{x^3}} + \color{maroon}{6{x^2}} }+{ \color{green}x – \color{red}{8{x^3}} – \color{maroon}{12{x^2}} }-{ \color{green}{6x} – \color{coral}1 } = {\color{blue}{8{x^4}} + \color{red}{4{x^3}} – \color{maroon}{6{x^2}} }-{ \color{green}{5x} – \color{coral}1.}$

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

${{y^{\left( n \right)}} = {\left( {\cos x} \right)^{\left( n \right)}} }={ \cos \left( {x + \frac{{n\pi }}{2}} \right).}$

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Problems 1-7
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Problems 8-22