### 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:

- A constant can be taken out of the differential sign:

\(d\left( {Cu} \right) = Cdu,\)

where \(C\) is a constant number. - The differential of the sum (difference) of two functions is equal to the sum (difference) of their differentials:

\({d\left( {u \pm v} \right) }={ du \pm dv.}\) - The differential of a constant is zero:

\(d\left( C \right) = 0.\) - The differential of the independent variable \(x\) is equal to its increment:

\(dx = \Delta x.\) - The differential of a linear function is equal to its increment:

\({d\left( {ax + b} \right) }\) \(= {\Delta \left( {ax + b} \right) }\) \(={ a\Delta x.}\) - Differential of the product of two functions:

\({d\left( {uv} \right) }={ du \cdot v + u \cdot dv.}\) - Differential of the quotient of two functions:

\({d\left( {\large\frac{u}{v}\normalsize} \right) }\) \(={ \large\frac{{du \cdot v – u \cdot dv}}{{{v^2}}}\normalsize.}\) - 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### Example 24

The function \(y\left( x \right)\) is defined by the parametric equations### 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

\[{dy = y’dx }={ 3{e^2}dx.}\]