(a) Use the method of first principles to determine the derivative of \( f(x)=\frac{6}{x} \) (b) Use an appropriate method of differentiation to determine the derivative of the following functions (sim- plify your answers as far as pnssible): (i) \( f(x)=\cos (\sqrt{\sin (\tan \pi x)}) \) (ii) \( p(t)-\frac{\cos (t)}{1-\sin (t)} \) (iii) \( g(x)-\ln \left(\frac{e^{x}}{1+e^{x}}\right) \)

To find the derivative of \( f(x)=\frac{6}{x} \) using first principles, we start with the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{6}{x+h} - \frac{6}{x}}{h} \] This simplifies to: \[ = \lim_{h \to 0} \frac{6(x - (x + h))}{h(x)(x + h)} = \lim_{h \to 0} \frac{-6h}{h(x)(x + h)} = \lim_{h \to 0} \frac{-6}{x(x + h)} = \frac{-6}{x^2} \] Thus, the derivative is \( f'(x) = -\frac{6}{x^2} \). Now, for the derivatives of the following functions: (i) For \( f(x) = \cos(\sqrt{\sin(\tan(\pi x))}) \), we use the chain rule. The derivative is: \[ f'(x) = -\sin(\sqrt{\sin(\tan(\pi x))}) \cdot \frac{1}{2\sqrt{\sin(\tan(\pi x))}} \cdot \cos(\tan(\pi x)) \cdot \pi \cdot \sec^2(\pi x) \] (ii) For \( p(t) = \frac{\cos(t)}{1 - \sin(t)} \), using the quotient rule: \[ p'(t) = \frac{-\sin(t)(1 - \sin(t)) - \cos(t)(-\cos(t)}{(1 - \sin(t))^2} = \frac{-\sin(t) + \cos^2(t)}{(1 - \sin(t))^2} \] (iii) For \( g(x)=\ln\left(\frac{e^x}{1+e^x}\right) \), we simplify first: \[ g(x) = x - \ln(1 + e^x) \] The derivative is: \[ g'(x) = 1 - \frac{e^x}{1 + e^x} = \frac{1}{1 + e^x} \]