Calculus

Applications of the Derivative

Applications of Derivative Logo

Linear Approximation

If the function \(y = f\left( x \right)\) is differentiable at a point \(a\), then the increment of this function when the independent variable changes by \(\Delta x\) is given by

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

where the first term \(A\Delta x\) is the differential of function, and the second term has a higher order of smallness with respect to \(\Delta x.\) The differential of function is denoted by \(dy\) and is linked with the derivative at the point \(a\) as follows:

\[dy = A\Delta x = f'\left( {a} \right)\Delta x.\]

Thus, the increment of the function \(\Delta y\) can be written as

\[\Delta y = dy + \omicron\left( {\Delta x} \right) = f'\left( {a} \right)\Delta x + \omicron\left( {\Delta x} \right).\]

For sufficiently small increments of the independent variable \(\Delta x\), one can neglect the "nonlinear" additive term \(\omicron\left( {\Delta x} \right).\) In this case, the following approximate equality is valid:

\[\Delta y \approx dy = f'\left( {a} \right)\Delta x.\]

Note that the absolute error of the approximation, i.e. the difference \(\Delta y - dy\) tends to zero as \(\Delta x \to 0:\)

\[\require{cancel} \lim\limits_{\Delta x \to 0} \left( {\Delta y - dy} \right) = \lim\limits_{\Delta x \to 0} \left[ {\cancel{dy} + \omicron\left( {\Delta x} \right) - \cancel{dy}} \right] = \lim\limits_{\Delta x \to 0} \omicron\left( {\Delta x} \right) = 0.\]

Moreover, the relative error also tends to zero as \(\Delta x \to 0:\)

\[\lim\limits_{\Delta x \to 0} \frac{{\Delta y - dy}}{{dy}} = \lim\limits_{\Delta x \to 0} \frac{{\omicron\left( {\Delta x} \right)}}{{f'\left( {a} \right)\Delta x}} = \frac{1}{{f'\left( {a} \right)}}\lim\limits_{\Delta x \to 0} \frac{{\omicron\left( {\Delta x} \right)}}{{\Delta x}} = 0,\]

since \(\omicron\left( {\Delta x} \right)\) corresponds to the term of the second and higher order of smallness with respect to \(\Delta x.\)

Thus, we can use the following formula for approximate calculations:

\[f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right).\]

where the function \(L\left( x \right)\) is called the linear approximation or linearization of \(f\left( x \right)\) at \(x = a.\)

linear approximation L(x) of a function f(x)
Figure 1.

Linear approximation is a good way to approximate values of \(f\left( x \right)\) as long as you stay close to the point \(x = a,\) but the farther you get from \(x = a,\) the worse your approximation.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Let \(f\left( x \right)\) be a differentiable function such that \(f\left( 3 \right) = 12,\) \(f^\prime\left( 3 \right) = - 2.\) Estimate the value of \(f\left( {3.5} \right)\) using the local approximation at \(a = 3.\)

Example 2

Let \(g\left( x \right)\) be a differentiable function such that \(g\left( { - 2} \right) = 0,\) \(g^\prime\left( { - 2} \right) = - 5.\) Find the value of \(g\left( { - 1.8} \right)\) using the local approximation at \(a = - 2.\)

Example 3

Find an approximate value for \(\sqrt[3]{{30}}.\)

Example 4

Estimate \(\sqrt[3]{9}\) using a linear approximation at \(a = 8.\)

Example 5

Calculate an approximate value of \(\sqrt {50}.\)

Example 6

Estimate \(\sqrt {3.9} \) using a linear approximation at \(a = 4.\)

Example 7

Calculate an approximate value for \(\sqrt[4]{{0,025}}.\)

Example 8

Find the linearization of the function \[f\left( x \right) = \sqrt[3]{{{x^2}}}\] at \(a = 27.\)

Example 9

Calculate \({\left( {8,2} \right)^{\frac{2}{3}}}.\)

Example 10

Derive the approximate formula \[{\left( {1 + \alpha } \right)^n} \approx 1 + n\alpha .\] Calculate the approximate value for \(\sqrt {1,02} .\)

Example 1.

Let \(f\left( x \right)\) be a differentiable function such that \(f\left( 3 \right) = 12,\) \(f^\prime\left( 3 \right) = - 2.\) Estimate the value of \(f\left( {3.5} \right)\) using the local approximation at \(a = 3.\)

Solution.

The linear approximation is given by the equation

\[f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).\]

We just need to plug in the known values and calculate the value of \(f\left( {3.5} \right):\)

\[L\left( x \right) = f\left( 3 \right) + f^\prime\left( 3 \right)\left( {x - 3} \right) = 12 - 2\left( {x - 3} \right) = 18 - 2x.\]

Then

\[f\left( {3.5} \right) \approx 18 - 2 \cdot 3.5 = 11.\]

Example 2.

Let \(g\left( x \right)\) be a differentiable function such that \(g\left( { - 2} \right) = 0,\) \(g^\prime\left( { - 2} \right) = - 5.\) Find the value of \(g\left( { - 1.8} \right)\) using the local approximation at \(a = - 2.\)

Solution.

First, we derive the linear approximation equation for the given function:

\[g\left( x \right) \approx L\left( x \right) = g\left( a \right) + g^\prime\left( a \right)\left( {x - a} \right),\]

where \(a = -2.\)

Hence

\[L\left( x \right) = 0 + \left( { - 5} \right)\left( {x - \left( { - 2} \right)} \right) = - 5x - 10.\]

This gives the following approximate value of \(g\left( { - 1.8} \right):\)

\[g\left( { - 1.8} \right) \approx - 5 \cdot \left( { - 1.8} \right) - 10 = - 1.\]

Example 3.

Find an approximate value for \(\sqrt[3]{{30}}.\)

Solution.

By the condition, \(x =30\). We take the start point \(a = 27.\) Then \(\Delta x = x - a = 30 - 27 = 3.\) The derivative of the function \(f\left( x \right) = \sqrt[3]{x}\) is given by

\[f'\left( x \right) = \left( {\sqrt[3]{x}} \right)^\prime = \left( {{x^{\frac{1}{3}}}} \right)^\prime = \frac{1}{3}{x^{ - \frac{2}{3}}} = \frac{1}{{3\sqrt[3]{{{x^2}}}}},\]

and its value at the point \(a\) is equal to

\[f'\left( {a} \right) = \frac{1}{{3\sqrt[3]{{{{27}^2}}}}} = \frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.\]

As a result, we get the following answer:

\[f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\Delta x,\;\; \Rightarrow \sqrt[3]{{30}} \approx \sqrt[3]{{27}} + \frac{1}{{27}} \cdot 3 = 3 + \frac{1}{9} = \frac{{28}}{9} \approx 3,111.\]

Example 4.

Estimate \(\sqrt[3]{9}\) using a linear approximation at \(a = 8.\)

Solution.

Let \(f\left( x \right) = \sqrt[3]{x}.\) The linear approximation at the point \(a = 8\) is given by

\[f\left( x \right) \approx L\left( x \right) = f\left( 8 \right) + f^\prime\left( 8 \right)\left( {x - 8} \right).\]

Find the derivative:

\[f^\prime\left( x \right) = \left( {\sqrt[3]{x}} \right)^\prime = \frac{1}{3}{x^{ - \frac{2}{3}}} = \frac{1}{{3\sqrt[3]{{{x^2}}}}}.\]

Compute the value of the derivative at \(a = 8:\)

\[f^\prime\left( 8 \right) = \frac{1}{{3\sqrt[3]{{{8^2}}}}} = \frac{1}{{12}}.\]

Substituting this, we get the function \(L\left( x \right)\) in the form

\[f\left( x \right) \approx L\left( x \right) = 2 + \frac{1}{{12}}\left( {x - 8} \right) = \frac{x}{{12}} + \frac{4}{3}.\]

Hence

\[\sqrt[3]{9} \approx L\left( 9 \right) = \frac{9}{{12}} + \frac{4}{3} = \frac{{9 + 16}}{{12}} = \frac{{25}}{{12}}.\]

Example 5.

Calculate an approximate value of \(\sqrt {50}.\)

Solution.

Consider the function \(f\left( x \right) = \sqrt x .\) In our case, it is necessary to find the value of this function at \(x = 50.\)

We choose \(a = 49\) and find the value of the derivative at this point:

\[f\left( x \right) = \sqrt x ,\;\; \Rightarrow f'\left( x \right) = \left( {\sqrt x } \right)^\prime = \frac{1}{{2\sqrt x }},\;\; \Rightarrow f'\left( {a = 49} \right) = \frac{1}{{2\sqrt {49} }} = \frac{1}{{14}}.\]

Using the formula

\[f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\Delta x,\]

we obtain:

\[\sqrt {50} \approx \sqrt {49} + \frac{1}{{14}} \cdot \left( {50 - 49} \right) = 7 + \frac{1}{{14}} = \frac{{99}}{{14}} \approx 7,071.\]

Example 6.

Estimate \(\sqrt {3.9} \) using a linear approximation at \(a = 4.\)

Solution.

We consider the linear approximation of \(f\left( x \right) = \sqrt x \) at the point \(a = 4.\)

The linearization of \(f\left( x \right)\) at \(a = 4\) is given by

\[f\left( x \right) \approx L\left( x \right) = f\left( 4 \right) + f^\prime\left( 4 \right)\left( {x - 4} \right).\]

Calculate the value of the function and its derivative at this point.

\[f^\prime\left( x \right) = \left( {\sqrt x } \right)^\prime = \frac{1}{{2\sqrt x }},\]
\[f\left( 4 \right) = \sqrt 4 = 2,\]
\[f^\prime\left( 4 \right) = \frac{1}{{2\sqrt 4 }} = \frac{1}{4}.\]

Plug this in the equation for \(L\left( x \right):\)

\[L\left( x \right) = 2 + \frac{1}{4}\left( {x - 4} \right) = 1 + \frac{x}{4}.\]

So, the answer is

\[\sqrt {3.9} \approx L\left( {3.9} \right) = 1 + \frac{{3.9}}{4} = 1.975\]

Example 7.

Calculate an approximate value for \(\sqrt[4]{{0,025}}.\)

Solution.

Here it is convenient to take the value \(a = 0,0256,\) since

\[f\left( {a} \right) = \sqrt[4]{{a}} = \sqrt[4]{{0,0256}} = 0,4.\]

Find the derivative of this function and its value at the point \(a:\)

\[f\left( x \right) = \sqrt[4]{x},\;\; \Rightarrow f'\left( x \right) = \left( {\sqrt[4]{x}} \right)^\prime = \left( {{x^{\frac{1}{4}}}} \right)^\prime = \frac{1}{4}{x^{ - \frac{3}{4}}} = \frac{1}{{4\sqrt[4]{{{x^3}}}}},\;\; \Rightarrow f'\left( {a = 0,0256} \right) = \frac{1}{{4\sqrt[4]{{0,{{0256}^3}}}}} = \frac{1}{{4{{\left( {\sqrt[4]{{0,0256}}} \right)}^3}}} = \frac{1}{{4 \cdot 0,{4^3}}} = \frac{1}{{4 \cdot 0,064}} = \frac{1}{{0,256}} \approx 3,9063.\]

Hence, we obtain the following approximate value of the function:

\[f\left( x \right) \approx L\left( x \right) = f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\;\; \Rightarrow \sqrt[4]{{0,025}} \approx 0,4 + 3,9063 \cdot \left( {0,025 - 0,0256} \right) = 0,4 + 3,9063 \cdot \left( { - 0,0006} \right) \approx 0,3977.\]

Example 8.

Find the linearization of the function \[f\left( x \right) = \sqrt[3]{{{x^2}}}\] at \(a = 27.\)

Solution.

We apply the formula

\[f\left( x \right) \approx L\left( x \right) = f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right),\]

where

\[f\left( a \right) = f\left( {27} \right) = \sqrt[3]{{{{27}^2}}} = 9.\]

Take the derivative using the power rule:

\[f^\prime\left( x \right) = \left( {\sqrt[3]{{{x^2}}}} \right)^\prime = \left( {{x^{\frac{2}{3}}}} \right)^\prime = \frac{2}{3}{x^{\frac{2}{3} - 1}} = \frac{2}{3}{x^{ - \frac{1}{3}}} = \frac{2}{{3\sqrt[3]{x}}}.\]

Then

\[f^\prime\left( a \right) = f^\prime\left( {27} \right) = \frac{2}{{3\sqrt[3]{{27}}}} = \frac{2}{9}.\]

Substitute this in the equation for \(L\left( x \right):\)

\[L\left( x \right) = 9 + \frac{2}{9}\left( {x - 27} \right) = 9 + \frac{2}{9}x - 6 = \frac{2}{9}x + 3.\]

Answer:

\[y = \frac{2}{9}x + 3.\]

Example 9.

Calculate \({\left( {8,2} \right)^{\frac{2}{3}}}.\)

Solution.

Here, obviously, \(f\left( x \right) = {x^{\frac{2}{3}}}\) and \(x = 8,2.\) Let \({a = 8}.\) Then

\[f\left( {a = 8} \right) = 8^{\frac{2}{3}} = \left( {\sqrt[3]{8}} \right)^2 = 2^2 = 4.\]

Find the derivative:

\[f'\left( x \right) = {\left( {{x^{\frac{2}{3}}}} \right)^\prime } = \frac{2}{3}{x^{ - \frac{1}{3}}} = \frac{2}{{3\sqrt[3]{x}}},\;\; \Rightarrow f'\left( {a = 8} \right) = \frac{2}{{3\sqrt[3]{8}}} = \frac{\cancel{2}}{{3 \cdot \cancel{2}}} = \frac{1}{3}.\]

The result is

\[f\left( x \right) \approx f\left( {a} \right) + f'\left( {a} \right)\left( {x - a} \right),\;\; \Rightarrow {\left( {8,2} \right)^{\frac{2}{3}}} \approx 4 + \frac{1}{3} \cdot \left( {8,2 - 8} \right) = 4 + \frac{1}{3} \cdot 0,2 \approx 4,067.\]

Example 10.

Derive the approximate formula \[{\left( {1 + \alpha } \right)^n} \approx 1 + n\alpha .\] Calculate the approximate value for \(\sqrt {1,02} .\)

Solution.

Consider the function \(f\left( x \right) = {x^n}.\) When the independent variable changes by \(\Delta x\), the increment of the function is given by

\[\Delta y = {\left( {x + \Delta x} \right)^n} - {x^n}.\]

If \(\Delta x\) is a small quantity, we can approximately assume that

\[\Delta y \approx dy = f'\left( x \right)\Delta x = {\left( {{x^n}} \right)^\prime }\Delta x = n{x^{n - 1}}\Delta x.\]

Consequently,

\[{\left( {x + \Delta x} \right)^n} \approx {x^n} + n{x^{n - 1}}\Delta x.\]

Suppose further that \(x = 1\) and \(\Delta x = \alpha.\) Then

\[{\left( {1 + \alpha } \right)^n} \approx {1^n} + n \cdot {1^{n - 1}} \cdot \alpha = 1 + n\alpha .\]

In particular,

\[\sqrt {1,02} = \sqrt {1 + 0,02} = {\left( {1 + 0,02} \right)^{\frac{1}{2}}} \approx 1 + \frac{1}{2} \cdot 0,02 = 1,01.\]

See more problems on Page 2.

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