Find the derivative of the function: $(x+\cos x)(x-\tan x)$
(N/A) Let $f(x) = (x+\cos x)(x-\tan x)$.
Using the product rule,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = (x+\cos x) \frac{d}{dx}(x-\tan x) + (x-\tan x) \frac{d}{dx}(x+\cos x)$
We know that $\frac{d}{dx}(x) = 1$,$\frac{d}{dx}(\cos x) = -\sin x$,and $\frac{d}{dx}(\tan x) = \sec^2 x$.
Substituting these derivatives:
$f'(x) = (x+\cos x)(1 - \sec^2 x) + (x-\tan x)(1 - \sin x)$
Since $1 - \sec^2 x = -\tan^2 x$,we have:
$f'(x) = (x+\cos x)(-\tan^2 x) + (x-\tan x)(1 - \sin x)$
$f'(x) = -\tan^2 x(x+\cos x) + (x-\tan x)(1 - \sin x)$
Using the product rule,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = (x+\cos x) \frac{d}{dx}(x-\tan x) + (x-\tan x) \frac{d}{dx}(x+\cos x)$
We know that $\frac{d}{dx}(x) = 1$,$\frac{d}{dx}(\cos x) = -\sin x$,and $\frac{d}{dx}(\tan x) = \sec^2 x$.
Substituting these derivatives:
$f'(x) = (x+\cos x)(1 - \sec^2 x) + (x-\tan x)(1 - \sin x)$
Since $1 - \sec^2 x = -\tan^2 x$,we have:
$f'(x) = (x+\cos x)(-\tan^2 x) + (x-\tan x)(1 - \sin x)$
$f'(x) = -\tan^2 x(x+\cos x) + (x-\tan x)(1 - \sin x)$
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