(i) Let f(x) = sin x cos x ... (1) Differentiating (1) with respect to x, we get f ′(x) = (sin x)′ cos x + (cos x)′ sin x = cosx · cosx + (– sin x) · sin x = $\cos^2 x - \sin^2 x$ ∴ f ′(x) = cos 2x. (ii) Let f(x) = sec x ⇒ f (x) = $\frac{1}{\cos x}$ ⇒ f (x) = $(\cos x)^{-1}$ ... (1) Differentiating (1) with respect to x, we get f ′(x) = – $(\cos x)^{-2}$ · (– sinx) ⇒ f' (x) = $\frac{\sin x}{\cos^2 x}$ ⇒ f' (x) = $\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}$ ∴ f ′(x) = tanx · secx. (iii) Let f (x) = 5 secx + 4 cosx ⇒ f (x) = $\frac{5}{\cos x} + 4 \cos x$ ... (1) Differentiating (1) with respect to x, we get f ′(x) = 5(–1) $(\cos x)^{-2}$ (– sin x) + 4 (– sin x) = $\frac{5\sin x}{\cos^2 x}$ - 4 sin x = $\frac{5\sin x}{\cos x} \cdot \frac{1}{\cos x}$ - 4 sin x ∴ f' (x) = 5 tan x . sec x - 4 sin x. (iv) Let f(x) = cosec x ⇒ f (x) = $\frac{1}{\sin x}$ ⇒ f (x) = $(\sin x)^{-1}$ ... (1) Differentiating (1) with respect to x, we get f ′(x) = (– 1) $(\sin x)^{-2}$ · cos x = $\frac{-\cos x}{\sin^2 x}$ = $\frac{-\cos x}{\sin x} \cdot \frac{1}{\sin x}$ ∴ f ′(x) = – cot x cosec x (v) Let f(x) = 3 cot x + 5 cosec x ⇒ f (x) = $\frac{3\cos x}{\sin x} + \frac{5}{\sin x}$ ... (1) Differentiating (1) with respect to x, we get f' (x) = $3\left[\frac{\sin x (\cos x)' - \cos x \cdot (\sin x)'}{\sin^2 x}\right]$ + $5\left[\frac{\sin x (1)' - 1 \cdot (\sin x)'}{\sin^2 x}\right]$ = $3\left[\frac{\sin x \cdot (-\sin x) - \cos x (\cos x)}{\sin^2 x}\right]$ + $5\left[\frac{0 - \cos x}{\sin^2 x}\right]$ = $3\left[\frac{-\sin^2 x - \cos^2 x}{\sin^2 x}\right]$ + $5\left[\frac{-\cos x}{\sin^2 x}\right]$ = $-3\left[\frac{\sin^2 x + \cos^2 x}{\sin^2 x}\right]$ - $5\left[\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}\right]$ = $-3\left[\frac{1}{\sin^2 x}\right]$ - 5 [cot x . cosec x] = – 3 $\csc^2$ x – 5 cot x · cosec x = – cosec x [3 cosec x + 5 cot x]. (vi) Let f(x) = 5 sin x – 6 cos x + 7 ... (1) Differentiating (1) with respect to x, we get f ′(x) = 5 cos x – 6 (– sin x) + 0 ∴ f ′(x) = 5 cos x + 6 sin x. (vii) Let f(x) = 2 tan x – 7 sec x ... (1) Differentiating (1) with respect to x, we get f ′(x) = 2 $\sec^2$ x – 7 sec x tan x