1. **Problem statement:** (i) Show that $\frac{d}{dx}(\sec x) = \sec x \tan x$ by writing $\sec x$ as $(\cos x)^{-1}$. (ii) Using this, find in terms of $\sec x$: (a) $\frac{d}{dx}(\sec x \tan x)$. (b) $\frac{d}{dx}[\ln(\sec x + \tan x)]$. 2. **Step (i): Differentiate $\sec x$ written as $(\cos x)^{-1}$** - Recall the chain rule: $\frac{d}{dx} [f(g(x))] = f'(g(x)) g'(x)$. - Write $\sec x = (\cos x)^{-1}$. - Differentiate: $$\frac{d}{dx}(\sec x) = \frac{d}{dx} (\cos x)^{-1} = -1 \cdot (\cos x)^{-2} \cdot (-\sin x) = \frac{\sin x}{\cos^2 x}.$$ - Rewrite using trigonometric identities: $$\frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x.$$ - Thus, $\frac{d}{dx}(\sec x) = \sec x \tan x$ as required. 3. **Step (ii)(a): Differentiate $\sec x \tan x$** - Use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$. - Let $u = \sec x$, $v = \tan x$. - From step (i), $u' = \sec x \tan x$. - Recall $\frac{d}{dx}(\tan x) = \sec^2 x$, so $v' = \sec^2 x$. - Apply product rule: $$\frac{d}{dx}(\sec x \tan x) = (\sec x \tan x)(\tan x) + (\sec x)(\sec^2 x) = \sec x \tan^2 x + \sec^3 x.$$ - Use identity $\tan^2 x = \sec^2 x - 1$: $$\sec x (\sec^2 x - 1) + \sec^3 x = \sec^3 x - \sec x + \sec^3 x = 2 \sec^3 x - \sec x = \sec x (2 \sec^2 x - 1).$$ 4. **Step (ii)(b): Differentiate $\ln(\sec x + \tan x)$** - Use chain rule: $\frac{d}{dx} \ln f(x) = \frac{f'(x)}{f(x)}$. - Let $f(x) = \sec x + \tan x$. - Differentiate $f(x)$: $$f'(x) = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) = \sec x \tan x + \sec^2 x.$$ - Factor $\sec x$: $$f'(x) = \sec x (\tan x + \sec x).$$ - Substitute back: $$\frac{d}{dx} \ln(\sec x + \tan x) = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} = \sec x.$$ **Final answers:** (i) $\frac{d}{dx}(\sec x) = \sec x \tan x$ (ii)(a) $\frac{d}{dx}(\sec x \tan x) = \sec x (2 \sec^2 x - 1)$ (ii)(b) $\frac{d}{dx}[\ln(\sec x + \tan x)] = \sec x$