Use differentials (or equivalently, a linear approximation ) to approximate ln(1.01) as follows: Let f(x)=ln(x) and find the equation of the tangent line to f(x) at a "nice" point near 1.01. Then use this to approximate ln(1.01).

Approximation=

Question

Use differentials (or equivalently, a linear approximation ) to approximate ln(1.01) as follows: Let f(x)=ln(x) and find the equation of the tangent line to f(x) at a "nice" point near 1.01. Then use this to approximate ln(1.01).

Approximation=

anonymous · Answer

$$f(x) = \ln(x)$$
$$\frac{df}{dx} = \frac{1}{x} $$
So, the equation of the tangent line at x = 1 has slope 
$$\frac{df}{dx} = \frac{1}{1} = 1$$
The equation of that line is
$$(y-0) = 1(x-1)\rightarrow y = x - 1 $$
So, near x = 1, we have that
$$f(x) \approx x - 1$$
so, 
$$\ln(1.01) \approx (1.01) - 1 = 0.01$$

anonymous · Answer

Thank you so much!