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 $$\frac{31}{30} - \sqrt[3]{1.1} \approx 1.033333 - 1.032280 = 0.001053$$