Question

How do you use a linear approximation or differentials to estimate `(2.001)^5`?

Answer 1

See below

Explanation:

We know that if `f(x)` is sufficiently differentiable then

`f(a+delta)approxf(a)+f^((1))(a)/(1!)delta + f^((2))(a)/(2!) delta^2+...+f^((n))(a)/(n!)delta^n + cdots`

Here `f^((i))` indicates the `i`-th derivative

Now making `a = 2` and `delta = 0.001`

`(2.001)^5 approx 2^5 + 5/(1!)cdot 2^4 cdot 0.001=32.08`