How do you write log to the base e?

Question

I was given a question, find $f'(1)$ of $f(x) = \ln \sqrt{2-x}$.

So I wrote

$$1/2 \ln (2-x)^{(-1/2)(-1)} = -1/2 \ln (2-x)^{-1/2}$$ $$= -(1/2\ln)/\sqrt{2-x}$$

But when I sub in $x = 1$ I get a SYNTAX error, I realised log base e cannot be put in my calculator. I don't know how to put this into my calculator, can anyone help? Thanks!!

Answer 1

Given, $$ f(x) = \ln \sqrt{2-x} $$

Use chain rule to differentiate.

$$ f^{'}(x) = \frac{1}{\sqrt{2-x}}\frac{1}{2 \sqrt{2-x}} (-1) $$

So, at $x=1$,

$$f^{'}(1) = \frac{-1}{2} $$

Answer 2

Using the chain rule, and since

$$\begin{cases}\left(\sqrt{2-x}\right)'=-\cfrac1{2\sqrt{2-x}}\\{}\\ \left(\log x\right)'=\cfrac1x\end{cases}\implies\left(\log\sqrt{2-x}\right)'=-\frac1{2\sqrt{2-x}}\cdot\frac1{\sqrt{2-x}}=-\frac1{2(2-x)}$$

for $\;x<2\;$, certainly.

Answer 3

$f(x)=\ln(\sqrt{2-x})=\frac{1}{2}\ln(2-x)$, hence $f'(x)=\frac{1}{2}\frac{1}{2-x}\cdot(-1)$.