# Calculus

## Differentiation of Functions # Differential of a Function

• ### Definition of the Differential of a Function

Consider a function $$y = f\left( x \right),$$ which is continuous in the interval $$\left[ {a,b} \right].$$ Suppose that at some point $${x_0} \in \left[ {a,b} \right]$$ the independent variable is incremented by $$\Delta x.$$ The increment of the function $$\Delta y$$ corresponding to the change of the independent variable $$\Delta x$$ is given by

${\Delta y = \Delta f\left( {{x_0}} \right) }={ f\left( {{x_0} + \Delta x} \right) – f\left( {{x_0}} \right).}$

For any differentiable function, the increment $$\Delta y$$ can be represented as a sum of two terms:

${\Delta y }={ A\Delta x + \omicron\left( {\Delta x} \right),}$

where the first term (called the principal part of the increment) is linearly dependent on the increment $$\Delta x,$$ and the second term has a higher order of smallness with respect to $$\Delta x.$$ The expression $$A\Delta x$$ is called the differential of function and is denoted by $$dy$$ or $$df\left( {{x_0}} \right).$$

Consider the idea of partition of the increment of of the function $$\Delta y$$ into two parts in the following simple example. Given a square with side $${x_0} = 1 \,\text{m}\,$$ (Figure $$1$$).

Its area is obviously equal to

${S_0} = x_0^2 = 1 \,\text{m}^2.$

If the side of the square is increased by $$\Delta x = 1\,\text{cm},$$ the exact value of the area of the square will be equal to

$S = {x^2} = {\left( {{x_0} + \Delta x} \right)^2} = {1,{01^2} }\kern0pt{\text{= } 1,0201 \,\text{m}^2,}\;$

that is the increment of the area $$\Delta S$$ is

${\Delta S = S – {S_0} }={ 1,0201 – 1 }={ 0,0201\,\text{m}^2 } = {201\,\text{cm}^2.}$

We now represent this increment $$\Delta S$$ as follows:

$\require{cancel} {\Delta S = S – {S_0} }={ {\left( {{x_0} + \Delta x} \right)^2} – x_0^2 } = {\cancel{x_0^2} + 2{x_0}\Delta x + {\left( {\Delta x} \right)^2} – \cancel{x_0^2} } = {2{x_0}\Delta x + {\left( {\Delta x} \right)^2} } = {A\Delta x + \omicron\left( {\Delta x} \right) } = {dy + o\left( {\Delta x} \right).}$

Thus, the increment $$\Delta S$$ consists of the principal part (the differential of the function), which is proportional to $$\Delta x$$ and is equal to

${dy = A\Delta x }={ 2{x_0}\Delta x }={ 2 \cdot 1 \cdot 0,01 }={ 0,02 \,\text{m}^2 }={ 200\,\text{cm}^2,}$

and the term of a higher order of smallness, which in turn is equal to

${\omicron\left( {\Delta x} \right) = {\left( {\Delta x} \right)^2} }={ {0,01^2} = 0,0001\,\text{m}^2 }={ 1\,\text{cm}^2.}$

In sum, both these terms comprise the full increment of the square area equal to $$200 + 1 = 201\,\text{cm}^2.$$

Note that in this example the coefficient $$A$$ is equal to the value of the derivative of $$S$$ at the point $${x_0}:$$

$A = 2{x_0}.$

It turns out that for any differentiable function, the following theorem is valid:

The coefficient $$A$$ in the principal part of the increment of a function at a point $${x_0}$$ is equal to the value of the derivative $$f’\left( {{x_0}} \right)$$ at this point, that is the increment $$\Delta y$$ is given by

${\Delta y = A\Delta x + \omicron\left( {\Delta x} \right) } = {f’\left( {{x_0}} \right)\Delta x + \omicron\left( {\Delta x} \right).}$

Dividing both sides of the equation by $$\Delta x \ne 0$$ gives

${\frac{{\Delta y}}{{\Delta x}} = A + \frac{{\omicron\left( {\Delta x} \right)}}{{\Delta x}} } = {f’\left( {{x_0}} \right) + \frac{{\omicron\left( {\Delta x} \right)}}{{\Delta x}}.}$

In the limit as $$\Delta x \to 0$$, we obtain the value of the derivative at the point $${x_0}:$$

${y’\left( {{x_0}} \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} } = {A }={ f’\left( {{x_0}} \right).}$

Here we took into account that for a small quantity $$\omicron\left( {\Delta x} \right)$$ of higher order of smallness than $$\Delta x,$$ the limit is equal to

$\lim\limits_{\Delta x \to 0} \frac{{\omicron\left( {\Delta x} \right)}}{{\Delta x}} = 0.$

Assuming that the differential of the independent variable $$dx$$ is equal to its increment $$\Delta x:$$

$dx = \Delta x,$

we obtain from the relationship

$dy = A\Delta x = y’dx$

that

$y’ = \frac{{dy}}{{dx}},$

i.e. the derivative of a function can be represented as the ratio of two differentials.

### Geometric Meaning of the Differential of a Function

Figure $$2$$ schematically shows splitting of the increment $$\Delta y$$ into the principal part $$A\Delta x$$ (the differential of function) and the term of a higher order of smallness $$\omicron\left( {\Delta x} \right).$$

The tangent $$MN$$ drawn to the curve of the function $$y = f\left( x \right)$$ at the point $$M,$$ as it is known, has the slope angle $$\alpha,$$ the tangent of which is equal to the derivative:

$\tan \alpha = f’\left( {{x_0}} \right).$

When the independent variable changes by $$\Delta x$$, the tangent increments by $$A\Delta x.$$ This linear increment formed by the tangent is just the differential of the function. The remaining part of the full increment $$\Delta y$$ (the segment $$N{M_1}$$) corresponds to the “nonlinear” additive of a higher order of smallness with respect to $$\Delta x.$$

### Properties of the Differential

Let $$u$$ and $$v$$ be functions of the variable $$x$$. The differential has the following properties:

1. A constant can be taken out of the differential sign:
$$d\left( {Cu} \right) = Cdu,$$
where $$C$$ is a constant number.
2. The differential of the sum (difference) of two functions is equal to the sum (difference) of their differentials:
$${d\left( {u \pm v} \right) }={ du \pm dv.}$$
3. The differential of a constant is zero:
$$d\left( C \right) = 0.$$
4. The differential of the independent variable $$x$$ is equal to its increment:
$$dx = \Delta x.$$
5. The differential of a linear function is equal to its increment:
$${d\left( {ax + b} \right) }$$ $$= {\Delta \left( {ax + b} \right) }$$ $$={ a\Delta x.}$$
6. Differential of the product of two functions:
$${d\left( {uv} \right) }={ du \cdot v + u \cdot dv.}$$
7. Differential of the quotient of two functions:
$${d\left( {\large\frac{u}{v}\normalsize} \right) }$$ $$={ \large\frac{{du \cdot v – u \cdot dv}}{{{v^2}}}\normalsize.}$$
8. The differential of a function is equal to the derivative of the function times the differential of the independent variable:
$${dy = df\left( x \right) }={ f’\left( x \right)dx.}$$

As you can see, the differential of the function $$dy$$ differs from the derivative only by the factor $$dx$$. For example,

${d\left( {{x^n}} \right) = n{x^{n – 1}}dx,\;\;\;}\kern0pt {d\left( {\ln x} \right) = \frac{{dx}}{x},\;\;\;}\kern0pt {d\left( {\sin x} \right) = \cos x dx}$

and so on.

### Form Invariance of the Differential

Consider a composition of two functions $$y = f\left( u \right)$$ and $$u = g\left( x \right).$$ Its derivative can be found by the chain rule:

${y’_x} = {y’_u} \cdot {u’_x},$

where the subindex denotes the variable of differentiation.

The differential of the “outer” function $$y = f\left( u \right)$$ can be written as

$dy = {y’_u}\,du.$

The differential of the “inner” function $$u = g\left( x \right)$$ can be represented in a similar manner:

$du = {u’_x}\,dx.$

If we substitute $$du$$ in the last formula, we obtain

${dy = {y’_u}\,du }={ {y’_u}{u’_x}\,dx.}$

Since $${y’_x} = {y’_u} \cdot {u’_x},$$ then

$dy = {y’_x}\,dx.$

It can be seen that in the case of a composite function, we get an expression for the differential in the same form as for a “simple” function. This property is called the form invariance of the differential.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Find the differential of the function $$y = \sin x – x\cos x.$$

### Example 2

Find the differential of the function $$y = \cot {\large\frac{{\pi x}}{4}\normalsize}$$ at the point $$x = 1.$$

### Example 3

Find the differential of the function $$y = 2{x^2} + 3x + 1$$ at the point $$x = 1$$ when $$dx = 0,1.$$

### Example 4

Calculate the increment and differential of the function $$y = {x^2} – x + 1$$ at the point $$x = 2$$ when $$dx = 1.$$

### Example 5

Determine the differential of the function $$y = {x^3} – 3{x^2} + 4x$$ at the point $${x_0} = 1$$ when $$dx = 0,1.$$

### Example 6

Calculate the differential of the function $$y = {x^5} – 7{x^3} + 5x$$ at the point $$x = 2$$ when $$dx = 0,2.$$

### Example 7

Use differential to approximate the change in $$y = {x^3} + {x^2}$$ as $$x$$ changes from $$1$$ to $$0,95.$$

### Example 8

Find the differential of the function $$y = {x^x}{e^{2x}}$$ at the point $$x = 1.$$

### Example 9

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

### Example 10

Find the differential of the function $$y = {\sin ^2}3x.$$

### Example 11

Find the differential of $$y = x\sin {\large\frac{{\pi x}}{2}\normalsize}$$ at the point $$x = {\large\frac{1}{2}\normalsize}$$ when $$dx = 0,01.$$

### Example 12

Find the differential of the function $$y = \sqrt {{x^3} + 4x}$$ at $$x = 2.$$

### Example 13

Use differential to approximate the change in $$y = \large{\frac{1}{{\sin x}}}\normalsize$$ as $$x$$ changes from $$\large{\frac{\pi }{4}}\normalsize$$ to $$\large{\frac{3\pi }{10}}\normalsize.$$

### Example 14

Calculate the increment and differential of the function $$y = {\large\frac{{x + 2}}{{x + 1}}\normalsize}$$ at the point $$x = 0$$ when $$\Delta x = 0,1.$$

### Example 15

Find the differential of the function $$y = \ln \tan 2x$$ at $$x = \large{\frac{\pi }{8}}\normalsize.$$

### Example 16

Calculate the differential of the function $$y = \large{\frac{{2x}}{{{x^2} – 3}}}\normalsize$$ at the point $$x = 3$$ when $$dx = – \large{\frac{3}{{15}}}\normalsize.$$

### Example 17

Find the differential of the function $$y = {\large\frac{1}{{\sqrt {{u^2} + {v^2}} }}\normalsize},$$ where $$u$$ and $$v$$ are differentiable functions of the variable $$x.$$

### Example 18

Find the differential of the function $$y = \arcsin {\large\frac{u}{v}\normalsize},$$ where $$u$$ and $$v$$ are differentiable functions of $$x.$$

### Example 19

The function $$y\left( x \right)$$ is given implicitly by the equation $${y^3} – 3xy + {x^3} = 3.$$ Find its differential at the point $$\left( {2,1} \right).$$

### Example 20

The function $$y\left( x \right)$$ is defined by the implicit equation $${x^2} – \sqrt y \,\ln y = 1.$$ Find its differential at the point $$\left( {1,1} \right).$$

### Example 21

Suppose that a force $$F$$ is determined by the law $$F = \large{\frac{1}{{{r^2}}}}\normalsize.$$ The distance $$r$$ is measured as $$r = 2 \pm 0,2.$$ Find the approximate value of the force $$F.$$

### Example 22

A body is moving through a liquid. The resistance force $$F$$ depends on the speed of the body as $$F = k{v^2},$$ where $$v$$ is the speed and $$k$$ is a constant. The speed $$v$$ was measured as $$v = 10 \pm 1.$$ Determine the approximate value of the resistance force $$F$$ if $$k = 1.$$

### Example 23

The function $$y\left( x \right)$$ is defined by the parametric equations
\left\{ \begin{aligned} x &= t^2 + t + 1 \\ y &= t^3 – 2t \end{aligned} \right..
Find the differential of the function at the point $$\left( {3, – 1} \right).$$

### Example 24

The function $$y\left( x \right)$$ is defined by the parametric equations
\left\{ \begin{aligned} x &= \left( {t + 2} \right){e^t} \\ y &= {e^{t + 1}} \end{aligned} \right..
Find the differential of the function at the point $$\left( {2, e} \right).$$

### Example 25

Given the composite function $$y = \ln u,\;u = \cos x.$$ Express the differential of $$y$$ in an invariant form.

### Example 1.

Find the differential of the function $$y = \sin x – x\cos x.$$

Solution.

Determine the derivative of the given function:

$\require{cancel} {y’ = {\left( {\sin x – x\cos x} \right)^\prime } } = {\cos x }-{ \left( {x’\cos x + x{{\left( {\cos x} \right)}^\prime }} \right) } = {\cos x }-{ \left( {\cos x + x\left( { – \sin x} \right)} \right) } = {\cancel{\cos x} – \cancel{\cos x} }+{ x\sin x }={ x\sin x.}$

The differential has the following form:

${dy = y’dx }={ x\sin x\,dx.}$

### Example 2.

Find the differential of the function $$y = \cot {\large\frac{{\pi x}}{4}\normalsize}$$ at the point $$x = 1.$$

Solution.

We find the derivative and calculate its value at the given point:

${y’ = \left( {\cot \frac{{\pi x}}{4}} \right)’ }={ – \frac{1}{{{{\sin }^2}\left( {\frac{{\pi x}}{4}} \right)}} \cdot \frac{\pi }{4} }={ – \frac{\pi }{{4{{\sin }^2}\left( {\frac{{\pi x}}{4}} \right)}},}$

$\Rightarrow {y’\left( 1 \right) = – \frac{\pi }{{4{{\sin }^2}\left( {\frac{\pi }{4}} \right)}} }={ – \frac{\pi }{{4{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}}} }={ – \frac{\pi }{2}.}$

Then

${dy = y’dx }={ – \frac{\pi }{2}dx.}$

### Example 3.

Find the differential of the function $$y = 2{x^2} + 3x + 1$$ at the point $$x = 1$$ when $$dx = 0,1.$$

Solution.

${dy = f’\left( x \right)dx }={ {\left( {2{x^2} + 3x + 1} \right)^\prime }dx } = {\left( {4x + 3} \right)dx.}$

Substituting the given values, we calculate the differential:

${dy = \left( {4 \cdot 1 + 3} \right) \cdot 0,1 }={ 0,7}$

### Example 4.

Calculate the increment and differential of the function $$y = {x^2} – x + 1$$ at the point $$x = 2$$ when $$dx = 1.$$

Solution.

Determine the increment of the function by the formula

${\Delta y }={ f\left( {x + \Delta x} \right) – f\left( x \right).}$

Since here $${x + \Delta x }={ 2 + 1 = 3,}$$ we have

${\Delta y = f\left( 3 \right) – f\left( 2 \right) } = {\left( {{3^2} – 3 + 1} \right) }-{ \left( {{2^2} – 2 + 1} \right) } = {7 – 3 = 4.}$

The differential (or the principal part of the increment) is given by

${dy = f’\left( x \right)\Delta x }={ {\left( {{x^2} – x + 1} \right)^\prime }\Delta x } = {\left( {2x – 1} \right)\Delta x } = {\left( {2 \cdot 2 – 1} \right) \cdot 1 }={ 3.}$

### Example 5.

Determine the differential of the function $$y = {x^3} – 3{x^2} + 4x$$ at the point $${x_0} = 1$$ when $$dx = 0,1.$$

Solution.

The derivative of the function $$y = f\left( x \right)$$ is given by

${f^\prime\left( x \right) = \left( {{x^3} – 3{x^2} + 4x} \right)^\prime }={ 3{x^2} – 6x + 4.}$

At the point $${x_0} = 1,$$ the derivative is equal to

${f^\prime\left( {{x_0}} \right) = f^\prime\left( 1 \right) }={ 3 \cdot {1^2} – 6 \cdot 1 + 4 }={ 1.}$

Substituting the values of $$f^\prime\left( {{x_0}} \right)$$ and $$dx,$$ we calculate the differential:

${dy = f^\prime\left( {{x_0}} \right)dx }={ 1 \cdot 0,1 }={ 0,1.}$

### Example 6.

Calculate the differential of the function $$y = {x^5} – 7{x^3} + 5x$$ at the point $$x = 2$$ when $$dx = 0,2.$$

Solution.

The derivative of the function is given by

${y^\prime = \left( {{x^5} – 7{x^3} + 5x} \right)^\prime }={ 5{x^4} – 21{x^2} + 5.}$

So at $$x = 2,$$ the derivative is equal to

${y^\prime\left( 2 \right) = 5 \cdot {2^4} – 21 \cdot {2^2} + 5 }={ 80 – 84 + 5 }={ 1.}$

The differential has the form

$dy = y^\prime dx.$

Substituting the values of $$y^\prime\left( 2 \right)$$ and $$dx,$$ we obtain:

${dy = y^\prime\left( 2 \right)dx }={ 1 \cdot 0,2 }={ 0,2.}$

### Example 7.

Use differential to approximate the change in $$y = {x^3} + {x^2}$$ as $$x$$ changes from $$1$$ to $$0,95.$$

Solution.

The differential $$dy$$ is defined by the formula

$dy = y^\prime dx = y^\prime\left( 1 \right)dx.$

Take the derivative

$y^\prime = \left( {{x^3} + {x^2}} \right)^\prime = 3{x^2} + 2x,$

so

${y^\prime\left( 1 \right) = 3 \cdot {1^2} + 2 \cdot 1 }={ 5.}$

Calculate the differential $$dx:$$

$dx = 0,95 – 1 = – 0,05.$

Hence

${dy = y^\prime\left( 1 \right)dx }={ 5 \cdot \left( { – 0,05} \right) }={ – 0,25.}$

The approximate value of the function at $$x = 0,95$$ is

${y\left( {0,95} \right) \approx y\left( 1 \right) + dy }={ \left( {{1^3} + {1^2}} \right) – 0,25 }={ 1,75.}$

### Example 8.

Find the differential of the function $$y = {x^x}{e^{2x}}$$ at the point $$x = 1.$$

Solution.

${y’ = {\left( {{x^x}{e^{2x}}} \right)^\prime } } = {{\left( {{x^x}} \right)^\prime }{e^{2x}} + {x^x}{\left( {{e^{2x}}} \right)^\prime }.}$

The derivative of the function $${x^x}$$ is

${{\left( {{x^x}} \right)^\prime } = {\left( {{e^{\ln x \cdot x}}} \right)^\prime } } = {{\left( {{e^{x\ln x}}} \right)^\prime } } = {{e^{x\ln x}}{\left( {x\ln x} \right)^\prime } } = {{x^x}\left( {1 \cdot \ln x + x \cdot \frac{1}{x}} \right) } = {{x^x}\left( {\ln x + 1} \right).}$

Hence, the derivative of the original function is given by

${y’ = {x^x}\left( {\ln x + 1} \right){e^{2x}} }+{ {x^x} \cdot 2{e^{2x}} } = {{x^x}{e^{2x}}\left( {\ln x + 1 + 2} \right) } = {{x^x}{e^{2x}}\left( {\ln x + 3} \right).}$

When $$x = 1,$$ we respectively have

${y’\left( 1 \right) }={ {1^1} \cdot {e^2} \cdot \left( {\ln 1 + 3} \right) }={ 3{e^2}.}$

Then the differential of the function at the given point is written as

${dy = y’dx }={ 3{e^2}dx.}$

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Problems 1-8
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Problems 9-25