# Thread: find f'(pi/4) of f(x) = [ln (cos x)]^2

1. ## find f'(pi/4) of f(x) = [ln (cos x)]^2

find f'(pi/4) of f(x) = [ln (cos x)]^2

I know f'(x) = -2 tan x ln (cos x)

two questions for f'(pi/4)

I know the answer is ln 2

so I know I compute -2 tan pi/4 ln (cos pi/4)

first question is tan pi/4 * cos pi/4 become 1/sqrt2. is there an easy way of remember such things?

next the problem then becomes -2 ln (2^-1/2)
which becomes ln 2. I am missing how it becomes ln 2. ??

find f'(pi/4) of f(x) = [ln (cos x)]^2

I know f'(x) = -2 tan x ln (cos x)

two questions for f'(pi/4)

I know the answer is ln 2

so I know I compute -2 tan pi/4 ln (cos pi/4)

first question is tan pi/4 * cos pi/4 become 1/sqrt2. is there an easy way of remember such things?

next the problem then becomes -2 ln (2^-1/2)
which becomes ln 2. I am missing how it becomes ln 2. ??
$\displaystyle f(x) = [\ln(\cos{x})]^2$

$\displaystyle f'(x) = 2\ln(\cos{x}) \cdot (-\tan{x})$

$\displaystyle f'\left(\frac{\pi}{4}\right) = 2\ln\left[\cos\left(\frac{\pi}{4}\right)\right] \cdot \left[-\tan\left(\frac{\pi}{4}\right)\right]$

$\displaystyle f'\left(\frac{\pi}{4}\right) = 2\ln\left(\frac{\sqrt{2}}{2}\right) \cdot \left(-1\right) = \ln\left(\frac{\sqrt{2}}{2}\right)^2 \cdot \left(-1\right) = \ln\left(\frac{1}{2}\right) \cdot \left(-1\right) = -\ln(2) \cdot (-1) = \ln(2)$

first question is tan pi/4 * cos pi/4 become 1/sqrt2. is there an easy way of remember such things?

Not this eaxctly, but you do have to remember that $\displaystyle \displaystyle\tan \frac{\pi}{4} = 1$ and $\displaystyle \displaystyle\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

next the problem then becomes -2 ln (2^-1/2)
which becomes ln 2. I am missing how it becomes ln 2. ??
$\displaystyle \displaystyle -2 \ln 2^{\frac{-1}{2}} = \frac{-1}{2}\times -2 \ln 2 = \ln 2$

4. is this from the log property loga (x^y) = y loga x

meaning ln 2^-1/2 = -1/2 ln 2?