Linear Approximations Using Differentials

If $f$ is a continuous function such that $f'(a) \ne 0$, then the tangent line to the curve $y = f(x)$ at $a$ provides a good approximation of the graph of $f$ near $x = a$. Recall that the tangent line at $a$ is given by

\[ y = f(a) + f'(a)(x-a) \] Instead of using the tangent line for the approximation, an equivalent formulation of the approximation would appeal directly to the definition of the derivative. That is, since \[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \] then for values $x$ close to $a$, we have that \[ f(x) \approx f'(a)(x - a) + f(a) \] If we subtract both sides by $f(a)$, then we arrive at \[ f(x) - (fa) \approx f'(a)(x-a) \] which can formulated in terms of the differentials \[ \Delta f \approx f'(a) \Delta x \]

Example

As an example, let's approximate the value of $\sqrt[3]{1.1}$.

We note that 1.1 is close to 1. Thus, we will make an approximation for $\sqrt[3]{1.1}$ based on the tangent line to the curve at $x=1$.

Since the derivative of $\sqrt[3]{x}$ is $\frac{1}{3 x^{2/3}}$ we have that the tangent line to curve at $x=1$ is \[ y = 1 + \frac{1}{3}(x - 1) \] Thus, substituting $1.1$ for $x$, we have as our approximation \[ y = 1 + \frac{1}{3} \frac{1}{10} = \frac{31}{30} \] It turns out that is approximation is quite good, as by calculator we can see that \[ \frac{31}{30} - \sqrt[3]{1.1} \approx 1.033333 - 1.032280 = 0.001053 \]