Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):$\frac{\sec x-1}{\sec x+1}$
Let $f(x)=\frac{\sec x-1}{\sec x+1}$
$f(x)=\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}=\frac{1-\cos x}{1+\cos x}$
By quotient rule,
$f^{\prime}(x)=\frac{(1+\cos x) \frac{d}{d x}(1-\cos x)-(1-\cos x) \frac{d}{d x}(1+\cos x)}{(1+\cos x)^{2}}$
$=\frac{(1+\cos x)(\sin x)-(1-\cos x)(-\sin x)}{(1+\cos x)^{2}}$
$=\frac{\sin x+\cos x \sin x+\sin x-\sin x \cos x}{(1+\cos x)^{2}}$
$=\frac{2 \sin x}{(1+\cos x)^{2}}$
$=\frac{2 \sin x}{\left(1+\frac{1}{\sec x}\right)^{2}}=\frac{2 \sin x}{\frac{(\sec x+1)^{2}}{\sec ^{2} x}}$
$=\frac{2 \sin x \sec ^{2} x}{(\sec x+1)^{2}}$
$=\frac{\frac{2 \sin x}{\cos x} \sec x}{(\sec x+1)^{2}}$
$=\frac{2 \sec x \tan x}{(\sec x+1)^{2}}$
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