# How do you graph y=ln(tan^2 x)?

Dec 23, 2017

see below

#### Explanation:

$f \left(x\right) = \ln \left({\tan}^{2} x\right)$
The Domain: ${\bigcup}_{k \in Z} \left(- \frac{\pi}{2} + k \pi , k \pi\right) \wedge {\bigcup}_{k \in Z} \left(k \pi , \frac{\pi}{2} + k \pi\right)$

$f \left(- x\right) = \ln {\left(\tan \left(- x\right)\right)}^{2}$
function tanx is odd: $\tan \left(- x\right) = - \tan x$
$\implies \ln \left({\left(- \tan x\right)}^{2}\right) \implies \ln \left[{\left(- 1\right)}^{2} \cdot {\left(\tan x\right)}^{2}\right] \implies \ln \left({\tan}^{2} x\right) = f \left(x\right)$

Function $\ln \left({\tan}^{2} x\right)$ is even
Has periodicity: $\pi$ so I will be graphing only the interval $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

$f ' \left(x\right) = \frac{1}{\tan} ^ 2 x \cdot 2 \tan x \cdot \frac{1}{\cos} ^ 2 x$

$f ' \left(x\right) = \frac{\cancel{{\cos}^{2} x}}{\sin} ^ 2 x \cdot 2 \tan x \cdot \frac{1}{\cancel{{\cos}^{2} x}}$

$f ' \left(x\right) = \frac{2 \tan x}{\sin} ^ 2 x$

$\tan x = 0 \Leftrightarrow x = 0$

$x \in \left(- \frac{\pi}{2} , 0\right) \Leftrightarrow f ' \left(x\right) < 0 \implies$f goes down

$x \in \left(0 , \frac{\pi}{2}\right) \Leftrightarrow f ' \left(x\right) > 0 \implies$f goes up

$f ' ' \left(x\right) = \frac{2 {\left(\sin x\right)}^{2} / {\left(\cos x\right)}^{2} - 2 \tan x \cdot 2 \sin x \cos x}{\sin x} ^ 4$

$f ' ' \left(x\right) = \frac{2 {\left(\sin x\right)}^{2} / {\left(\cos x\right)}^{2} - 2 \left(\sin \frac{x}{\cancel{\cos}} x\right) \cdot 2 \sin x \cancel{\cos} x}{\sin x} ^ 4$

$f ' ' \left(x\right) = \frac{\frac{2}{\cos x} ^ 2 - 2 \cdot 2}{\sin x} ^ 2$

$f ' ' \left(x\right) = \frac{2 \left(1 - 2 {\cos}^{2} x\right)}{{\left(\cos x\right)}^{2} {\left(\sin x\right)}^{2}}$

${\cos}^{2} x = \frac{1}{2} \implies \cos x = \pm \frac{\sqrt{2}}{2}$

$\cos x = \pm \frac{\sqrt{2}}{2} \implies x = \pm \frac{\pi}{4}$

$x \in \left(- \frac{\pi}{2} , - \frac{\pi}{4}\right) \Leftrightarrow f ' ' \left(x\right) > 0 \implies$f is convex U

$x \in \left(- \frac{\pi}{4} , \frac{\pi}{4}\right) \Leftrightarrow f ' ' \left(x\right) < 0 \implies$f is concave $\cap$

$x \in \left(\frac{\pi}{4} , \frac{\pi}{2}\right) \Leftrightarrow f ' ' \left(x\right) > 0 \implies$f is convex U

$f \left(\frac{\pi}{4}\right) = \ln \left(1\right) = 0$

$L i {m}_{x \rightarrow {0}^{+}} \ln \left({0}^{+}\right) \approx \ln \left({10}^{}\right) \approx - 100000 \left(\ln 10\right) \approx - \infty$

Now we have everything we need to make a graph.

The Range: $\mathbb{R}$