(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) \( g(x)-\ln \left(\frac{e^{x}}{1+e^{x}}\right) \)

Certainly! Let's tackle each part of your question step by step. --- ### **Problem (a):** **Use the method of first principles to determine the derivative of \( f(x) = \frac{6}{x} \).** **Solution:** The derivative of a function \( f(x) \) using first principles (the definition of the derivative) is given by: \[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \] Let's apply this to \( f(x) = \frac{6}{x} \): 1. **Compute \( f(x + h) \):** \[ f(x + h) = \frac{6}{x + h} \] 2. **Find the difference \( f(x + h) - f(x) \):** \[ f(x + h) - f(x) = \frac{6}{x + h} - \frac{6}{x} = 6\left( \frac{1}{x + h} - \frac{1}{x} \right) \] 3. **Simplify the difference:** \[ 6\left( \frac{1}{x + h} - \frac{1}{x} \right) = 6 \left( \frac{x - (x + h)}{x(x + h)} \right) = 6 \left( \frac{-h}{x(x + h)} \right) = \frac{-6h}{x(x + h)} \] 4. **Divide by \( h \) and take the limit as \( h \to 0 \):** \[ f'(x) = \lim_{h \to 0} \frac{-6h}{x(x + h)h} = \lim_{h \to 0} \frac{-6}{x(x + h)} = \frac{-6}{x^2} \] **Answer:** \[ f'(x) = -\frac{6}{x^2} \] --- ### **Problem (b)(i):** **Differentiate \( f(x) = \cos\left( \sqrt{\sin(\tan(\pi x))} \right) \).** **Solution:** To find the derivative of a composite function, we'll use the chain rule multiple times. Let's break it down step by step. 1. **Identify the innermost function and work outward:** - Let \( u = \tan(\pi x) \) - Then \( v = \sin(u) = \sin(\tan(\pi x)) \) - Next, \( w = \sqrt{v} = \sqrt{\sin(\tan(\pi x))} \) - Finally, \( f(x) = \cos(w) = \cos\left( \sqrt{\sin(\tan(\pi x))} \right) \) 2. **Differentiate each layer using the chain rule:** - \( \frac{du}{dx} = \pi \sec^2(\pi x) \) (Derivative of \( \tan(\pi x) \) is \( \pi \sec^2(\pi x) \)) - \( \frac{dv}{du} = \cos(u) = \cos(\tan(\pi x)) \) - \( \frac{dw}{dv} = \frac{1}{2\sqrt{v}} = \frac{1}{2\sqrt{\sin(\tan(\pi x))}} \) - \( \frac{df}{dw} = -\sin(w) = -\sin\left( \sqrt{\sin(\tan(\pi x))} \right) \) 3. **Apply the chain rule by multiplying the derivatives:** \[ f'(x) = \frac{df}{dw} \cdot \frac{dw}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} \] Substituting the derivatives: \[ f'(x) = \left[ -\sin\left( \sqrt{\sin(\tan(\pi x))} \right) \right] \cdot \left[ \frac{1}{2\sqrt{\sin(\tan(\pi x))}} \right] \cdot \left[ \cos(\tan(\pi x)) \right] \cdot \left[ \pi \sec^2(\pi x) \right] \] 4. **Simplify the expression:** \[ f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^2(\pi x)}{2\sqrt{\sin(\tan(\pi x))}} \] **Answer:** \[ f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^2(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}} \] --- ### **Problem (b)(ii):** **Differentiate \( p(t) = \frac{\cos(t)}{1 - \sin(t)} \).** **Solution:** We'll use the quotient rule for differentiation, which states: \[ \left( \frac{N}{D} \right)' = \frac{N' D - N D'}{D^2} \] For \( p(t) = \frac{\cos(t)}{1 - \sin(t)} \): 1. **Identify \( N(t) \) and \( D(t) \):** - \( N(t) = \cos(t) \) - \( D(t) = 1 - \sin(t) \) 2. **Find the derivatives \( N'(t) \) and \( D'(t) \):** - \( N'(t) = -\sin(t) \) - \( D'(t) = -\cos(t) \) 3. **Apply the quotient rule:** \[ p'(t) = \frac{(-\sin(t))(1 - \sin(t)) - \cos(t)(-\cos(t))}{(1 - \sin(t))^2} \] 4. **Simplify the numerator:** \[ -\sin(t)(1 - \sin(t)) + \cos^2(t) = -\sin(t) + \sin^2(t) + \cos^2(t) \] Recall that \( \sin^2(t) + \cos^2(t) = 1 \), so: \[ -\sin(t) + 1 = 1 - \sin(t) \] 5. **Final expression for the derivative:** \[ p'(t) = \frac{1 - \sin(t)}{(1 - \sin(t))^2} = \frac{1}{1 - \sin(t)} \] **Answer:** \[ p'(t) = \frac{1}{1 - \sin(t)} \] --- ### **Problem (b)(iii):** **Differentiate \( g(x) = \ln\left( \frac{e^x}{1 + e^x} \right) \).** **Solution:** First, let's simplify the expression before differentiating. 1. **Simplify \( g(x) \):** \[ g(x) = \ln\left( \frac{e^x}{1 + e^x} \right) = \ln(e^x) - \ln(1 + e^x) = x - \ln(1 + e^x) \] 2. **Differentiate \( g(x) \):** \[ g'(x) = \frac{d}{dx} \left( x - \ln(1 + e^x) \right) = 1 - \frac{d}{dx} \ln(1 + e^x) \] 3. **Differentiate \( \ln(1 + e^x) \):** Using the chain rule: \[ \frac{d}{dx} \ln(1 + e^x) = \frac{e^x}{1 + e^x} \] 4. **Combine the results:** \[ g'(x) = 1 - \frac{e^x}{1 + e^x} = \frac{(1 + e^x) - e^x}{1 + e^x} = \frac{1}{1 + e^x} \] **Answer:** \[ g'(x) = \frac{1}{1 + e^{x}} \] --- ### **Summary of Answers:** - **(a)** \( f'(x) = -\dfrac{6}{x^{2}} \) - **(b)(i)** \[ f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^{2}(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}} \] - **(b)(ii)** \[ p'(t) = \frac{1}{1 - \sin(t)} \] - **(b)(iii)** \[ g'(x) = \frac{1}{1 + e^{x}} \]