1. **Problem Statement:** Find the derivatives of the following functions: a) $y = \sin(\cos(\tan x))$ b) $y = \sin x + \cos(\sin x)$ c) $y = \ln\left(\frac{\sin x - 1}{\sin x + 1}\right)$ 2. **Formulas and Rules:** - Derivative of $\sin u$ is $\cos u \cdot u'$ - Derivative of $\cos u$ is $-\sin u \cdot u'$ - Derivative of $\tan x$ is $\sec^2 x$ - Derivative of $\ln u$ is $\frac{u'}{u}$ - Derivative of sum is sum of derivatives - Derivative of quotient $\frac{f}{g}$ is $\frac{f'g - fg'}{g^2}$ 3. **Solution:** a) $y = \sin(\cos(\tan x))$ Let $u = \cos(\tan x)$, then $y = \sin u$ First, find $u' = -\sin(\tan x) \cdot \sec^2 x$ Then, $y' = \cos(u) \cdot u' = \cos(\cos(\tan x)) \cdot \left(-\sin(\tan x) \cdot \sec^2 x\right)$ So, $$y' = -\cos(\cos(\tan x)) \sin(\tan x) \sec^2 x$$ b) $y = \sin x + \cos(\sin x)$ Derivative of $\sin x$ is $\cos x$ For $\cos(\sin x)$, let $v = \sin x$, then derivative is $-\sin(v) \cdot v' = -\sin(\sin x) \cdot \cos x$ So, $$y' = \cos x - \sin(\sin x) \cos x = \cos x \left(1 - \sin(\sin x)\right)$$ c) $y = \ln\left(\frac{\sin x - 1}{\sin x + 1}\right)$ Let $f = \sin x - 1$, $g = \sin x + 1$ Then, $$y' = \frac{\frac{d}{dx}(f/g)}{f/g} = \frac{\frac{f'g - fg'}{g^2}}{f/g} = \frac{f'g - fg'}{fg}$$ Calculate derivatives: $f' = \cos x$ $g' = \cos x$ Substitute: $$y' = \frac{\cos x (\sin x + 1) - (\sin x - 1) \cos x}{(\sin x - 1)(\sin x + 1)} = \frac{\cos x \sin x + \cos x - \cos x \sin x + \cos x}{\sin^2 x - 1} = \frac{2 \cos x}{\sin^2 x - 1}$$ 4. **Final answers:** a) $$y' = -\cos(\cos(\tan x)) \sin(\tan x) \sec^2 x$$ b) $$y' = \cos x \left(1 - \sin(\sin x)\right)$$ c) $$y' = \frac{2 \cos x}{\sin^2 x - 1}$$