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 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 \(\sqrt[3]{9}\) 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 \(\sqrt[\large 4\normalsize]{{0,025}}.\)

    Example 8

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

    Example 9

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

    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 \(\sqrt[3]{9}\) 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 \(\sqrt[\large 4\normalsize]{{0,025}}.\)

    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 \(a:\)

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

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

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

    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:

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

    Page 1
    Problems 1-8
    Page 2
    Problems 9-25