Question

given ln(10)=2.3, use differentials to approximate ln(10.2)

Answer 1

This is the same procedure as with using a linear approximation. To approximate a function at some value \(x\) near a known \(a\), you have
\[\begin{align*}
f(x)&\approx f'(a)(x-a)+f(a)
\end{align*}\]
\(f(x)=\ln x~~\implies ~~f'(x)=\dfrac{1}{x}\). In this case, \(a=10\) and \(x=10.2\).
\[\begin{align*}
\ln10.2&\approx \frac{1}{10}(10.2-10)+2.3\\
&=0.02+2.3\\
&=2.32
\end{align*}\]