Calculus

Applications of the Derivative

Approximation by Differentials

Page 1
Problems 1-3
Page 2
Problems 4-15

If the function \(y = f\left( x \right)\) is differentiable at \({x_0}\), 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 \({x_0}\) as follows:
\[dy = A\Delta x = f’\left( {{x_0}} \right)\Delta x.\] Thus, the increment of the function \(\Delta y\) can be written as
\[
{\Delta y = dy + \omicron\left( {\Delta x} \right) }
= {f’\left( {{x_0}} \right)\Delta x + \omicron\left( {\Delta x} \right).}
\] 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 folowing approximate equality is valid:
\[\Delta y \approx dy = f’\left( {{x_0}} \right)\Delta x.\] 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 \(\Delta x \to 0:\)
\[
{\lim\limits_{\Delta x \to 0} \frac{{\Delta y – dy}}{{dy}} }
= {\lim\limits_{\Delta x \to 0} \frac{{\omicron\left( {\Delta x} \right)}}{{f’\left( {{x_0}} \right)\Delta x}} }
= {\frac{1}{{f’\left( {{x_0}} \right)}}\lim\limits_{\Delta x \to 0} \frac{{\omicron\left( {\Delta x} \right)}}{{\Delta x}} = 0,}
\] 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:
\[f\left( x \right) \approx f\left( {{x_0}} \right) + f’\left( {{x_0}} \right)\Delta x,\] where \(\Delta x = x – {x_0}\) and \(\Delta y = f\left( x \right) – f\left( {{x_0}} \right).\)

Solved Problems

Click on problem description to see solution.

 Example 1

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

 Example 2

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

 Example 3

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

 Example 4

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

 Example 5

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

 Example 6

Derive the approximate formula
\[{\sqrt {{a^2} + h} \approx a + \frac{h}{{2a}}\;\;}\kern-0.3pt{\left( {a \gt 0} \right).}\] Using this identity calculate approximately the value of \(\sqrt {150} .\)

 Example 7

Derive the approximate formula
\[{\sqrt[\large n\normalsize]{{{a^n} + h}} \approx a + \frac{h}{{n{a^{n – 1}}}}\;\;}\kern-0.3pt{\left( {a \gt 0} \right).}\] Using this formula, calculate \(\sqrt[\large 8\normalsize]{{250}}.\)

 Example 8

Find an approximate value for \(\cos 46^\circ.\)

 Example 9

Find an approximate value for \(\sin 179^\circ.\)

 Example 10

Find an approximate value of \(\ln 20.\)

 Example 11

Calculate \({e^{0,1}}.\)

 Example 12

Find an approximate value for \(\arccos 0,51.\)

 Example 13

Find an approximate value for \(\arctan 0,95.\)

 Example 14

Find the approximate value of the function \(f\left( x \right) = \sqrt {{x^2} + 3x} \) at \(x = 1,02.\)

 Example 15

Find the approximate value of the function \(f\left( x \right) = \sqrt {5x – 1} \) at \(x = 1,99.\)

Example 1.

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

Solution.

By the condition, \(x =30\). We take the start point \({x_0} = 27.\) Then \(\Delta x = x – {x_0} = 30 – 27 = 3.\) The derivative of the function \(f\left( x \right) = \sqrt[\large 3\normalsize]{x}\) is given by
\[
{f’\left( x \right) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } }
= {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } }
= {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}} }
= {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}},}
\] and its value at the point \({x_0}\) is equal to
\[
{f’\left( {{x_0}} \right) = \frac{1}{{3\sqrt[\large 3\normalsize]{{{{27}^2}}}}} }
= {\frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.}
\] As a result, we get the following answer:
\[
{f\left( x \right) \approx f\left( {{x_0}} \right) + f’\left( {{x_0}} \right)\Delta x,\;\;}\Rightarrow
{\sqrt[\large 3\normalsize]{{30}} \approx \sqrt[\large 3\normalsize]{{27}} + \frac{1}{{27}} \cdot 3 }
= {3 + \frac{1}{9} }
= {\frac{{28}}{9} \approx 3,111.}
\]

Example 2.

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 \({x_0} = 49\) and find the value of the derivative at this point:
\[
{f\left( x \right) = \sqrt x ,\;\;}\Rightarrow
{f’\left( x \right) = {\left( {\sqrt x } \right)^\prime } = \frac{1}{{2\sqrt x }},\;\;}\Rightarrow
{f’\left( {{x_0} = 49} \right) = \frac{1}{{2\sqrt {49} }} = \frac{1}{{14}}.}
\] Using the formula
\[f\left( x \right) \approx f\left( {{x_0}} \right) + f’\left( {{x_0}} \right)\Delta x,\] 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 3.

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

Solution.

Here it is convenient to take the value \({x_0} = 0,0256,\) since
\[
{f\left( {{x_0}} \right) = \sqrt[\large 4\normalsize]{{{x_0}}} }
= {\sqrt[\large 4\normalsize]{{0,0256}} = 0,4.}
\] Find the derivative of this function and its value at the point \({x_0}:\)
\[
{f\left( x \right) = \sqrt[4]{x},\;\;}\Rightarrow
{f’\left( x \right) = {\left( {\sqrt[\large 4\normalsize]{x}} \right)^\prime } }
= {{\left( {{x^{\large\frac{1}{4}\normalsize}}} \right)^\prime } }
= {\frac{1}{4}{x^{ – \large\frac{3}{4}\normalsize}} }
= {\frac{1}{{4\sqrt[\large 4\normalsize]{{{x^3}}}}},\;\;}\Rightarrow
{f’\left( {{x_0} = 0,0256} \right) }
= {\frac{1}{{4\sqrt[\large 4\normalsize]{{0,{{0256}^3}}}}} }
= {\frac{1}{{4{{\left( {\sqrt[\large 4\normalsize]{{0,0256}}} \right)}^3}}} }
= {\frac{1}{{4 \cdot 0,{4^3}}} }
= {\frac{1}{{4 \cdot 0,064}} }
= {\frac{1}{{0,256}} \approx 3,9063.}
\] Hence, we obtain the following approximate value of the function:
\[
{f\left( x \right) \approx f\left( {{x_0}} \right) + f’\left( {{x_0}} \right)\left( {x – {x_0}} \right),\;\;}\Rightarrow
{\sqrt[4]{{0,025}} \approx 0,4 + 3,9063 \cdot \left( {0,025 – 0,0256} \right) }
= {0,4 + 3,9063 \cdot \left( { – 0,0006} \right)} \approx {0,3977.}
\]

Page 1
Problems 1-3
Page 2
Problems 4-15