Estimating (2.001)^5 with Differentials and Linear Approximations

In our posts so far on the topic of differentials, we have seen that the differential dy is the rise or fall of of the tangent line L(x) of a function f(x) while Δy is the rise or fall of the curve. This is illustrated in Figure 1 below.

Figure 1.
Created with: www.desmos.com/calculator

As we can see from Figure 1, as dx becomes smaller, the difference between Δy and dy approaches 0.

Now...

...is the y-coordinate of the point Q in Figure 1.

The tangent line L(x) is the linear approximation of f(x) near x = a. By this definition, the linear approximation of the point Q can also be written in terms of differentials as...

Let's use this approximation to estimate the number...

For this, let's consider the function...

with x = a = 2 and dx = 0.001.

With differentials, we have...

And now, by linear approximation, we have...

A scientific calculator output of the result is...

Thus the error in our result is within 0.01%. In general, a very good approximation indeed for the given dx.

Now, as pointless of an exercise as this seems, this form of linear approximation can be very useful. In this case we were able to calculate the exact value of (2.001)⁵ using a calculator. But even calculators are not able to calculate exact values of some funtions, and this is where this use of linear approximations can come in very handy.

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Here's a list of posts created so far on the subject of Linear Approximations and Differentials:

1. Linear Approximations Part 1 - Interpolating between Empirical Data
2. Linear Approximations Part 2 - Estimating values of f(x) = √(x+2) near x = 2
3. Linear Approximations Part 3 - why sin(x) ≅ x near 0
4. The Geometric Meaning of Differentials
5. Differentials: Comparing dy and Δy for y = x^2 at x = 1 and Δx = 0.5
6. Estimating (2.001)^5 with Differentials and Linear Approximations

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