# 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

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

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:
$d\left( {\frac{u}{v}} \right) = \frac{{du \cdot v - u \cdot dv}}{{{v^2}}}.$
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,

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

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 or tap 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 $y = \cot {\frac{{\pi x}}{4}}$ 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 1.

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

Solution.

Determine the derivative of the given function:

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

The differential has the following form:

$dy = y'dx = x\sin x\,dx.$

### Example 2.

Find the differential of the function $y = \cot {\frac{{\pi x}}{4}}$ 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

$dy = y'dx = 3{e^2}dx.$

See more problems on Page 2.