# Calculus

## Applications of the Derivative # 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 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 $$\sqrt[\large 3\normalsize]{{30}}.$$

### Example 4

Estimate $$\sqrt{9}$$ 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 $$\sqrt[\large 4\normalsize]{{0,025}}.$$

### Example 8

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

### Example 9

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

### Example 10

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

### Example 11

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 12

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 13

Find an approximate value for $$\cos 46^\circ.$$

### Example 14

Find the linearization of the function $$f\left( x \right) = {x^2} + 2\cos x$$ at $$a = 0.$$

### Example 15

Find an approximate value for $$\sin 179^\circ.$$

### Example 16

Find the linearization of the natural logarithm $$f\left( x \right) = \ln x$$ at $$x = 1.$$

### Example 17

Find an approximate value of $$\ln 20.$$

### Example 18

Calculate $${e^{0,1}}.$$

### Example 19

Find an approximate value for $$\arccos 0,51.$$

### Example 20

Find an approximate value for $$\arctan 0,95.$$

### Example 21

Given the function $$f\left( x \right) = \large{\frac{3}{{{x^2}}}}\normalsize.$$ Use linear approximation at $$a = 2$$ to estimate the value of $$f\left( {1.99} \right).$$

### Example 22

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

### Example 23

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

### Example 24

Find the linear approximation to $$f\left( x \right) = \large{\frac{{{x^2} + 1}}{{x – 2}}}\normalsize$$ near $$a = 3.$$

### Example 25

The function $$f\left( x \right) = – {x^2} + 100x$$ is linearized at a point $$P\left( {a,f\left( a \right)} \right).$$ The linear approximation line $$L\left( x \right)$$ passes through $$P$$ and intersects the $$x-$$axis at $$x = 180.$$
1. Find the coordinates $$\left( {a,f\left( a \right)} \right)$$ of the point $$P.$$
2. Write the linearization equation $$L\left( x \right).$$

### 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 $$\sqrt[\large 3\normalsize]{{30}}.$$

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 $$f\left( x \right) = \sqrt[\large 3\normalsize]{x}$$ 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 $$a$$ is equal to

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

As a result, we get the following answer:

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

### Example 4.

Estimate $$\sqrt{9}$$ using a linear approximation at $$a = 8.$$

Solution.

Let $$f\left( x \right) = \sqrt{x}.$$ 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:

${f^\prime\left( x \right) = \left( {\sqrt{x}} \right)^\prime }={ \frac{1}{3}{x^{ – \frac{2}{3}}} }={ \frac{1}{{3\sqrt{{{x^2}}}}}.}$

Compute the value of the derivative at $$a = 8:$$

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

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

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

### 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 $$L\left( x \right):$$

${L\left( x \right) = 2 + \frac{1}{4}\left( {x – 4} \right) }={ 1 + \frac{x}{4}.}$

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

### Example 7.

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

Solution.

Here it is convenient to take the value $$a = 0,0256,$$ since

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

Find the derivative of this function and its value at the point $$a:$$

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

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

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

### Example 8.

Find the linearization of the function $$f\left( x \right) = \sqrt{{{x^2}}}$$ 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

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

Take the derivative using the power rule:

${f^\prime\left( x \right) = \left( {\sqrt{{{x^2}}}} \right)^\prime }={ \left( {{x^{\frac{2}{3}}}} \right)^\prime }={ \frac{2}{3}{x^{\frac{2}{3} – 1}} }={ \frac{2}{3}{x^{ – \frac{1}{3}}} }={ \frac{2}{{3\sqrt{x}}}.}$

Then

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

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

${L\left( x \right) = 9 + \frac{2}{9}\left( {x – 27} \right) }={ 9 + \frac{2}{9}x – 6 }={ \frac{2}{9}x + 3.}$

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