Is the linear approximation of the product of two functions the same as the product of the linear approximations of the two functions

calculuslinear approximation

I have just seen a lecture about linear approximation, in which it was established that when $x$ is near $0$:

  1. The linear approximation of $e^x$ is $1+x$

  2. The linear approximation of $(1+x)^r$ is $1+rx$

Afterwards, the lecturer explained how to find—using 1 and 2—the linear approximation of:

$e^{-3x} \times (1+x)^{-1/2}$.

His next line is:

$(1-3x)(1-0.5x)$

So it seems that he was acting under the assumption that the linear approximation of the product of two functions is the product of the linear approximations of the functions.

Is that true in general for any order of approximations? Maybe just for linear ones? Maybe just when x is near zero? Maybe only when the approximation is linear and x near 0?

Here is a link to the lecture. The comments were made at 26:30.

Best Answer

This is commonly used, and that is because the difference between the two is negligible.

In general, if we have a function $f(x)$ approximated by $f(x_0)+f'(x_0)(x-x_0)$ and another function $g(x)$ approximated the same then you can do some algebra to multiply the two approximations in a part and multiply the two functions then approximate in the other part and compare the results, it won't be the same but the difference would be pretty negligible as long as $x$ is close to $x_0$ so you can use the two methods interchangeably, to be exact the difference would be $f'(x_0)g'(x_0)(x-x_0)^2$.

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