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# 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