If the function y = f (x) is differentiable at a point a, then the increment of this function when the independent variable changes by Δx is given by
\[\Delta y = A\Delta x + \omicron\left( {\Delta x} \right),\]
where the first term AΔx is the differential of function, and the second term has a higher order of smallness with respect to Δx. The differential of function is denoted by Δy 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 Δy 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 \(x = a.\)
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
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 \(f\left( {3.5} \right):\)
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:
\[g\left( x \right) \approx L\left( x \right) = g\left( a \right) + g^\prime\left( a \right)\left( {x - a} \right),\]
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
