# Question #ca7f8

May 20, 2015

So you don't know the derivative of $\ln \left(x\right)$, right?

(I assume the function is $f \left(x\right) = \frac{1}{2} x \ln \left({x}^{2}\right)$, otherwise $- 1 \notin D \left(f\right)$, so $f \left(x\right) = x \ln \left(\left\mid x \right\mid\right)$, but please in the future write LaTeX-ly)

Let's write the definition:

${\lim}_{h \to 0} \frac{\ln \left(x + h\right) - \ln \left(x\right)}{h} = {\lim}_{h \to 0} \ln \frac{1 + \frac{h}{x}}{h}$

You know ${\lim}_{a \to 0} \ln \frac{1 + a}{a} = 1$, so

${\lim}_{h \to 0} \ln \frac{1 + \frac{h}{x}}{h} = {\lim}_{h \to 0} \frac{\frac{h}{x}}{h} = \frac{1}{x}$

So, by Leibniz

$f ' \left(x\right) = \ln \left(\left\mid x \right\mid\right) + x \frac{1}{x} = \ln \left(\left\mid x \right\mid\right) + 1$

So the equation of the tangent line is

$y - {y}_{0} = f ' \left({x}_{0}\right) \left(x - {x}_{0}\right)$

Which means

$y = \left(1 + \ln \left(1\right)\right) \left(x + 1\right) = x + 1$

graph{1/2xln(x^2) [-10, 10, -5, 5]}
graph{y=x+1 [-10, 10, -5, 5]}

May 20, 2015

For $f \left(x\right) = \frac{1}{2} x \ln {x}^{2}$,

to find the slope of the tangent line at $\left(- 1 , 0\right)$, we need $f ' \left(- 1\right)$.

Usually, I like to use the fact that $\ln {x}^{r} = r \ln x$, but that assumes that $x > 0$, which we will not have for the point $\left(- 1 , 0\right)$, so leave it as is.

To find $f ' \left(x\right)$ we'll use the product rule:

$f ' \left(x\right) = \frac{1}{2} \ln {x}^{2} + \frac{1}{2} x \left[\frac{1}{x} ^ 2 \cdot 2 x\right]$

$\left(\text{Note: } \frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}\right)$

So, $f ' \left(x\right) = \frac{1}{2} \ln {x}^{2} + 1$ .

And at the point of interest: $f ' \left(- 1\right) = \frac{1}{2} \ln 1 + 1 = 0 + 1 = 1$

I assume you can find the equation of the line through $\left(- 1 , 0\right)$ with slope $m = 1$.

We could have used $\ln {x}^{2} = 2 \ln \left\mid x \right\mid$ together with
$\frac{d}{\mathrm{dx}} \ln \left\mid x \right\mid = \frac{1}{x}$ to get
$\frac{d}{\mathrm{dx}} \left(\ln {x}^{2}\right) = \frac{d}{\mathrm{dx}} \left(2 \ln \left\mid x \right\mid\right) = 2 \frac{d}{\mathrm{dx}} \left(\ln \left\mid x \right\mid\right) = \frac{2}{x}$