Differential of a Function

Definition of the Differential of a Function

Consider a function y = f (x), which is continuous in the interval [a, b]. Suppose that at some point x₀ ∈ [a, b] the independent variable is incremented by Δx. The increment of the function Δy corresponding to the change of the independent variable Δx is given by

For any differentiable function, the increment Δy can be represented as a sum of two terms:

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

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

that is the increment of the area \(\Delta S\) is

We now represent this increment \(\Delta S\) as follows:

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

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

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 \({x_0}:\)

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

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

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

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

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

we obtain from the relationship

that

so 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:

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,

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:

where the subindex denotes the variable of differentiation.

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

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

If we substitute \(du\) in the last formula, we obtain

Since \({y'_x} = {y'_u} \cdot {u'_x},\) then

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

Solution.

Determine the derivative of the given function:

The differential has the following form:

Solution.

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

Then

Solution.

Substituting the given values, we calculate the differential:

Solution.

Determine the increment of the function by the formula

Since here \(x + \Delta x = 2 + 1 = 3,\) we have

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

Solution.

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

At the point \({x_0} = 1,\) the derivative is equal to

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

Solution.

The derivative of the function is given by

So at \(x = 2,\) the derivative is equal to

The differential has the form

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

Solution.

The differential \(dy\) is defined by the formula

Take the derivative

so

Calculate the differential \(dx:\)

Hence

The approximate value of the function at \(x = 0,95\) is

Solution.

The derivative of the function \({x^x}\) is

Hence, the derivative of the original function is given by

When \(x = 1,\) we respectively have

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