What is the derivative of $- cos x \cdot ln (sec x + tan x)$ $-cosx*ln(secx+tanx)$ ?

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What is the derivative of $- cos x \cdot ln (sec x + tan x)$ $-cosx*ln(secx+tanx)$ ?

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Answer 1 · 2015-05-09T02:24:56

Just do the product rule, followed by the chain rule (on only the $ln (u)$ $ln(u)$ ). Recall that $\frac{d}{d x} [cos x] = - sin x$ $d/(dx)[cosx] = -sinx$ , and that $\frac{d}{d x} [ln (u (x))] = \frac{1}{u (x)} \frac{d u (x)}{d x} = \frac{1}{u (x)} (\frac{d}{d x} [u (x)])$ $d/(dx)[ln(u(x))] = 1/(u(x))(du(x))/(dx) = 1/(u(x))((d)/(dx)[u(x)])$

So:
$\frac{d}{d x} [- cos x \cdot ln (sec x + tan x)]$ $d/(dx)[-cosx*ln(secx+tanx)]$

$= [- cos x \cdot (\frac{1}{sec x + tan x}) \cdot (sec x tan x + {sec}^{2} x)] + [ln (sec x + tan x) sin x]$ $= [-cosx*(1/(secx+tanx))*(secxtanx+sec^2x)]+[ln(secx+tanx)sinx]$

Move the cosine in:
$= [(- \frac{cos x}{sec x + tan x}) \cdot (sec x tan x + {sec}^{2} x)] + [ln (sec x + tan x) sin x]$ $= [(-cosx/(secx+tanx))*(secxtanx+sec^2x)]+[ln(secx+tanx)sinx]$

Factor and cancel:
$= [(-cosx/cancel(secx+tanx))*{secx(cancel(tanx+secx))}]+[ln(secx+tanx)sinx]$

Notice that $secx = 1/cosx$ :
$= [-cosx*secx]+[ln(secx+tanx)sinx]$

And you get:
$= -1+sinxln(secx+tanx)$

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What is the derivative of $- cos x \cdot ln (sec x + tan x)$ $-cosx*ln(secx+tanx)$ ?

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