Question

use differential to approximate ln(0.91)

Answer 1

\[
f(x) = \ln(x)
\]

Answer 2

In this case, we can let \(x_0=1\), since it is simple to evaluate: \[
f(x_0) = f(1) = \ln(1) = 0
\]

Answer 3

ok i get that part what do I do with the 0.91?

Answer 4

We use \[
f(x) - f(x_0) = \Delta f \approx df = f'(x_0) dx = f'(x_0) \Delta x = f'(x_0)(x-x_0)
\]

Answer 5

In short, \[
f(x) = f'(x_0)(x-x_0)
\]In our case: \[
\ln(0.91) = \frac{d(\ln(x))}{dx}\Bigg|^{x=1} (1-0.91)
\]

Answer 6

Except change \(=\) to \(\approx\).

Answer 7

1-0.91 =0.09

Answer 8

so now do i subtact o.o9 from ln(0.910

Answer 9

subtract*

Answer 10

Hold on let me fix real quick: \[
f(x) \approx f'(x_0)(x-x_0)+f(x_0)
\]In our case: \[
\ln(0.91) \approx \frac{d(\ln(x))}{dx}\Bigg|^{x=1} (1-0.91) + \ln(1)
\]

Answer 11

Now you need to differentiate and evaluate at \(x=1\).

Answer 12

im confused

Answer 13

You need to find \(f'(1)\).

Answer 14

ln(1) = 0

Answer 15

No, first differentiate.

Answer 16

ok hold on

Answer 17

0

Answer 18

Okay, so \(\ln(1) = 0\), that is true... we have: \[
\ln(0.91) = 0.09\cdot f'(1) +0
\]

Answer 19

\(\approx \)

Answer 20

0.09

Answer 21

So you got \(f'(1) = 1\)?

Answer 22

no im confused on this

Answer 23

we can move on tho