# How do you find the derivatives of y=lnabs(cosx)?

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#### Explanation

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1
Jan 16, 2017

$- \tan x , x \ne$ an odd multiple of $\frac{\pi}{2}$.

#### Explanation:

To make y real, $x \ne$ an odd multiple of $\frac{\pi}{2}$'

Applying function of function of function rule,

$y ' = \frac{1}{|} \cos x | | \cos x | '$

$= \frac{1}{|} \cos x | \left(1\right) \left(\cos x\right) ' = - \sin \frac{x}{|} \cos x | = - \sin \frac{x}{\cos} x = - \tan x$,

when $\cos x > 0 \to x \in o p e n {Q}_{1} \bigcup {Q}_{4}$ and

$= \frac{1}{|} \cos x | \left(- 1\right) \left(\cos x\right) ' = \sin \frac{x}{|} \cos x | = \sin \frac{x}{- \cos x} = \tan x$,

when $\cos x < 0 \to x \in o p e n {Q}_{2} \bigcup {Q}_{3}$.

In brief,

$y ' = - \tan x , x \ne$ an odd multiple of $\frac{\pi}{2}$.

Then teach the underlying concepts
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#### Explanation

Explain in detail...

#### Explanation:

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1
Jan 16, 2017

$\frac{d}{\mathrm{dx}} \left(\ln \left\mid \cos x \right\mid\right) = - \tan x$

#### Explanation:

We note that the function:

$f \left(x\right) = \ln \left\mid \cos x \right\mid$

is defined in the intervals:

$x = \left(- \frac{\pi}{2} + k \pi , \frac{\pi}{2} + k \pi\right)$ with $k \in \mathbb{Z}$

where $\cos x \ne 0$

In the central interval $x \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ we have $\cos x > 0$, so that:

$\ln \left\mid \cos x \right\mid = \ln \left(\cos x\right)$

$\frac{d}{\mathrm{dx}} \ln \left\mid \cos x \right\mid = \frac{d}{\mathrm{dx}} \ln \left(\cos x\right) = - \sin \frac{x}{\cos} x = - \tan x$

In any other interval we have:

$x \in \left(- \frac{\pi}{2} + k \pi , \frac{\pi}{2} + k \pi\right) \implies y = \left(x - k \pi\right) \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

but:

$\left\mid \cos x \right\mid = \left\mid \cos \left(y + k \pi\right) \right\mid = \left\mid {\left(- 1\right)}^{k} \cos y \right\mid = \left\mid \cos y \right\mid$

so that $f \left(x\right)$ is periodic of period $\pi$ and $f ' \left(x\right)$ must also be necessarily periodic of period $\pi$

In conclusion:

$\frac{d}{\mathrm{dx}} \left(\ln \left\mid \cos x \right\mid\right) = - \tan x$

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