# Calculus

## Differentiation of Functions # Higher-Order Differentials

• ### Concept of Higher-Order Differentials

We consider a function $$y = f\left( x \right)$$, which is differentiable in the interval $$\left( {a,b} \right).$$ The first-order differential of the function at the point $$x \in \left( {a,b} \right)$$ is defined by the formula

$dy = f’\left( x \right)dx.$

It can be seen that the differential $$dy$$ depends on two quantities – the variable $$x$$ (through the derivative $$y = f’\left( x \right)$$) and the differential of the independent variable $$dx.$$

Let us fix the increment $$dx$$, i.e. we assume that $$dx$$ is constant. Then the differential $$dy$$ becomes a function only of the variable $$x,$$ for which we can also define the differential by taking the same differential $$dx$$ as the increment $$\Delta x.$$ As a result, we obtain the second differential or differential of the second order, which is denoted as $${d^2}y$$ or $${d^2}f\left( x \right).$$ Thus, by definition:

${{d^2}y = d\left( {dy} \right) } = {d\left[ {f’\left( x \right)dx} \right] } = {df’\left( x \right)dx } = {f^{\prime\prime}\left( x \right)dxdx } = {f^{\prime\prime}\left( x \right){\left( {dx} \right)^2}.}$

Its is commonly denoted $${\left( {dx} \right)^2} = d{x^2}.$$ Therefore, we get:

${d^2}y = f^{\prime\prime}\left( x \right)d{x^2}.$

In the same way, we can establish that the third differential or differential of the third order has the form

${d^3}y = f^{\prime\prime\prime}\left( x \right)d{x^3}.$

In the general case, the differential of an arbitrary order $$n$$ is given by

${d^n}y = {f^{\left( n \right)}}\left( x \right)d{x^n},$

which can be rigorously proved by mathematical induction. This formula leads in particular to the following expression for the $$n$$th order derivative:

${f^{\left( n \right)}}\left( x \right) = \frac{{{d^n}y}}{{d{x^n}}}.$

Note that for the linear function $$y = ax + b,$$ the second and subsequent higher-order differentials are zero. Indeed,

${{d^2}\left( {ax + b} \right) }= {\left( {ax + b} \right)^{\prime\prime}} d{x^2} = {0 \cdot d{x^2} = 0,} \ldots , {{d^n}\left( {ax + b} \right) = 0.}$

In this case, it is obvious that

${{d^n}x = 0\;\;}\kern-0.3pt{\text{for}\;\;n \gt 1.}$

### Properties of Higher-Order Differentials

Let the functions $$u$$ and $$v$$ have the $$n$$th order derivatives. Then the following properties are valid:

• $${{d^n}\left( {\alpha u + \beta v} \right) }$$ $$={ \alpha {d^n}u + \beta {d^n}v;}$$
• $${{d^n}\left( {uv} \right) }$$ $$={ \sum\limits_{i = 0}^n {C_n^i{d^{n – i}}u{d^i}v}.}$$

The last equality follows directly from the Leibniz formula.

### Higher Order Differential of a Composite Function

Consider now the composition of two functions such that $$y = f\left( u \right)$$ and $$u = g\left( x \right).$$ In this case, $$y$$ is a composite function of the independent variable $$x:$$

$y = f\left( {g\left( x \right)} \right).$

The first differential of $$y$$ can be written as

${dy = {\left[ {f\left( {g\left( x \right)} \right)} \right]^\prime }dx } = {f’\left( {g\left( x \right)} \right)g’\left( x \right)dx.}$

Compute the second differential $${d^2}y$$ (assuming $$dx$$ is constant by definition). Using the product rule, we obtain:

${{d^2}y }={ {\left[ {f’\left( {g\left( x \right)} \right)g’\left( x \right)} \right]^\prime }d{x^2} } = {\left[ {f^{\prime\prime}\left( {g\left( x \right)} \right){{\left( {g’\left( x \right)} \right)}^2} }\right.}+{\left.{ f’\left( {g\left( x \right)} \right)g^{\prime\prime}\left( x \right)} \right]d{x^2} } = {f^{\prime\prime}\left( {g\left( x \right)} \right){\left( {g’\left( x \right)dx} \right)^2} }+{ f’\left( {g\left( x \right)} \right)g^{\prime\prime}\left( x \right)d{x^2}.}$

Take into account that

${g’\left( x \right)dx = du\;\;\text{and}\;\;\;}\kern-0.3pt{g^{\prime\prime}\left( x \right)d{x^2} = {d^2}u.}$

Consequently,

${{d^2}y = f^{\prime\prime}\left( u \right)d{u^2} }+{ f’\left( u \right){d^2}u}$

or in short form:

${{d^2}y }={ y^{\prime\prime}d{u^2} + y'{d^2}u.}$

In the same way, we can obtain the expression for the third order differential of a composite function:

${{d^3}y = f^{\prime\prime\prime}\left( u \right)d{u^3} }+{ 3f^{\prime\prime}\left( u \right)du{d^2}u }+{ f^\prime\left( u \right){d^3}u.}$

It follows from the above that the higher order differentials $${d^2}y,{d^3}y, \ldots ,{d^n}y$$ are generally not invariant.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Find the differential $${d^4}y$$ of the function $$y = {x^4}.$$

### Example 2

Find the differential $${d^4}y$$ of the function $$y = {x^5}.$$

### Example 3

Given the function $$y = \sqrt{x}$$. Find $${d^2}y.$$

### Example 4

Find the differential $${d^5}y$$ of the function $$y = \sin 2x.$$

### Example 5

Find the second differential of the function $$y = {x^2}\cos 2x.$$

### Example 6

Find $${d^3}y$$ of the function $$y = x\ln \large{\frac{1}{x}}\normalsize.$$

### Example 7

Find the $$4$$th order differential of the function $$y = \sqrt {2x + 1}.$$

### Example 8

Find the differential $${d^3}y$$ of the function $$y = \large{\frac{1}{{\sqrt x }}}\normalsize.$$

### Example 9

Find $${d^3}y$$ of the function $$y = {x^2}\ln x.$$

### Example 10

Find the $$n$$th-order differential of the function $$y = {2^x}.$$

### Example 11

Find the differential $${d^3}y$$ of the function $$y = {x^2}{e^x}.$$

### Example 12

Find the differential $${d^5}y$$ of the function $$y = \ln {x^2}.$$

### Example 13

Find the $$5$$th order differential of the function $$y = x\cos x.$$

### Example 14

Find the differential $${d^3}y$$ of the function $$y = x\sinh x.$$

### Example 15

Find the differential $${d^8}y$$ of the function $$y = {e^x}\sin x.$$

### Example 16

Given the function $$y = \sin x\cosh x.$$ Find $${d^3}y.$$

### Example 17

Find the second differential $${d^2}y$$ of the astroid defined by the equation ${x^{\large\frac{2}{3}\normalsize}} + {y^{\large\frac{2}{3}\normalsize}} = {R^{\large\frac{2}{3}\normalsize}}.$

### Example 18

The function is given in parametric form by the equations \left\{ \begin{aligned} x &= {t^2} + t – 1 \\ y &= {t^3} – 2t \end{aligned} \right.. Find the second-order differential $${d^2}y.$$

### Example 19

The equation of the upper semicircle centered at the origin with radius $$R$$ in parametric form is given by ${x = R\cos t,\;\;}\kern-0.3pt{y = R\sin t,\;\;\;}\kern-0.3pt {\text{where}\;\;0 \lt t \lt \pi .}$ Find the third differential $${d^3}y$$ of this function.

### Example 20

Find the second order differential of the function $$y = {u^3},$$ where $$u = u\left( x \right).$$

### Example 21

Find the second order differential of the function $$y = {e^{2u}},$$ where $$u = u\left( x \right).$$

### Example 1.

Find the differential $${d^4}y$$ of the function $$y = {x^4}.$$

Solution.

Determine the $$4$$th derivative:

$y^\prime = \left( {{x^4}} \right)^\prime = 4{x^3};$

$y^{\prime\prime} = \left( {4{x^3}} \right)^\prime = 4 \cdot 3{x^2} = 12{x^2};$

$y^{\prime\prime\prime} = \left( {12{x^2}} \right)^\prime = 12 \cdot 2x = 24x;$

${y^{\left( 4 \right)}} = \left( {24x} \right)^\prime = 24.$

Then the differential $${d^4}y$$ is given by

${d^4}y = 24d{x^4}.$

### Example 2.

Find the differential $${d^4}y$$ of the function $$y = {x^5}.$$

Solution.

The $$4$$th order differential is given by

${d^4}y = {f^{\left( 4 \right)}}\left( x \right)d{x^4} = {\left( {{x^5}} \right)^{\left( 4 \right)}}d{x^4}.$

We find the fourth derivative of this function by successive differentiation:

${{\left( {{x^5}} \right)’} = 5{x^4},\;\;\;}\kern-0.3pt {{\left( {{x^5}} \right)^{\prime\prime}} = {\left( {5{x^4}} \right)’} = 20{x^3},\;\;\;}\kern-0.3pt {{\left( {{x^5}} \right)^{\prime\prime\prime}} = {\left( {20{x^3}} \right)’} = 60{x^2},\;\;\;}\kern-0.3pt {{\left( {{x^5}} \right)^{\left( 4 \right)}} = {\left( {60{x^2}} \right)’} = 120x.}$

Hence,

${d^4}y = 120x\,d{x^4}.$

### Example 3.

Given the function $$y = \sqrt{x}$$. Find $${d^2}y.$$

Solution.

We calculate the first and second derivatives of the given function:

${y’ = f’\left( x \right) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } = {\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{\large\frac{1}{3}\normalsize – 1}} = {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}},}\;\;\;}\kern-0.3pt {y^{\prime\prime} = {f^{\prime\prime}\left( x \right)} = {\frac{1}{3} \cdot \left( { – \frac{2}{3}} \right){x^{ – \large\frac{5}{3}\normalsize}} }= { – \frac{2}{{9\sqrt[\large 3\normalsize]{{{x^5}}}}}.}}$

Then the second-order differential is written as

${{d^2}y = f^{\prime\prime}\left( x \right)d{x^2} }={ – \frac{{2d{x^2}}}{{9\sqrt[\large 3\normalsize]{{{x^5}}}}}.}$

### Example 4.

Find the differential $${d^5}y$$ of the function $$y = \sin 2x.$$

Solution.

It is known that the $$n$$th-order derivative of the sine function has the form

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

One can show that the $$n$$th-order derivative of the function $$y = \sin 2x$$ is given by

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

Hence, the $$5$$th-order derivative is written as

${{y^{\left( 5 \right)}} = {\left( {\sin 2x} \right)^{\left( 5 \right)}} }={ {2^5}\sin \left( {2x + \frac{{5\pi }}{2}} \right) }={ 32\sin \left( {2x + 2\pi + \frac{\pi }{2}} \right) }={ 32\sin \left( {2x + \frac{\pi }{2}} \right) }={ 32\cos 2x.}$

This yields

${d^5}y = 32\cos 2x\,d{x^5}.$

### Example 5.

Find the second differential of the function $$y = {x^2}\cos 2x.$$

Solution.

Determine the second derivative of this function:

${y’ = {\left( {{x^2}\cos 2x} \right)^\prime } } = {{\left( {{x^2}} \right)^\prime }\cos 2x }+{ {x^2}{\left( {\cos 2x} \right)^\prime } } = {2x\cos 2x }+{ {x^2} \cdot \left( { – 2\sin 2x} \right) }={ 2x\cos 2x }-{ 2{x^2}\sin 2x,}$

${y^{\prime\prime} }={ {\left( {2x\cos 2x – 2{x^2}\sin 2x} \right)^\prime } } = {2{\left( {x\cos 2x – {x^2}\sin 2x} \right)^\prime } } = {2\left[ {x’\cos 2x + x{{\left( {\cos 2x} \right)}^\prime } }\right.}\kern0pt{-\left.{ {{\left( {{x^2}} \right)}^\prime }\sin 2x – {x^2}{{\left( {\sin 2x} \right)}^\prime }} \right] } = {2\left[ {\cos 2x – 2x\sin 2x }\right.}-{\left.{ 2x\sin 2x – 2{x^2}\cos 2x} \right] } = {\left( {2 – 2{x^2}} \right)\cos 2x }-{ 4x\sin 2x.}$

Then the second-order differential is written in the form:

${ {d^2}y = y^{\prime\prime}d{x^2} } = {\left[ {\left( {2 – 2{x^2}} \right)\cos 2x }-{ 4x\sin 2x} \right]d{x^2}}$

### Example 6.

Find $${d^3}y$$ of the function $$y = x\ln \large{\frac{1}{x}}\normalsize.$$

Solution.

The third order differential is given by

${{d^3}y = {y^{\prime\prime\prime}\left( x \right)}d{x^3}.}$

We differentiate the given function successively:

${y^\prime = \left( {x\ln \frac{1}{x}} \right)^\prime }={ x \cdot \left( {\frac{1}{{\frac{1}{x}}}} \right) \cdot \left( {\frac{1}{x}} \right)^\prime + 1 \cdot \ln \frac{1}{x} }={ {x^2} \cdot \left( { – \frac{1}{{{x^2}}}} \right) + \ln \frac{1}{x} }={ \ln \frac{1}{x} – 1;}$

${y^{\prime\prime} = \left( {\ln \frac{1}{x} – 1} \right)^\prime }={ \left( {\frac{1}{{\frac{1}{x}}}} \right) \cdot \left( {\frac{1}{x}} \right)^\prime }={ x \cdot \left( { – \frac{1}{{{x^2}}}} \right) }={ – \frac{1}{x};}$

${y^{\prime\prime\prime} = \left( { – \frac{1}{x}} \right)^\prime }={ \frac{1}{{{x^2}}}.}$

Hence

${d^3}y = y^{\prime\prime\prime}\left( x \right)d{x^3} = {\frac{{d{x^3}}}{{{x^2}}}.}$

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Problems 7-21