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# 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 ) $\large\frac{\sec x - 1}{\sec x +1}$

Let $f(x) = \large\frac{\sec x - 1}{\sec x +1}$
$f(x) = \large\frac{\Large\frac{1}{\cos x}-1}{\Large\frac{1}{\cos x}+1}$$=\large\frac{1- \cos x}{1+ \cos x} By quotient rule, f'(x) = \large\frac{(1+ \cos x ) \large\frac{d}{dx}(1-\cos x )-(1-\cos x )\large\frac{d}{dx}(1+\cos x )}{(1+\cos x)^2} = \large\frac{(1+\cos x )(\sin x)- (1-\cos x)(-\sin x)}{(1+ \cos x)^2} = \large\frac{\sin x+\cos x \sin x+ \sin x-\sin x \cos x }{(1+ \cos x)^2} = \large\frac{2\sin x}{(1+ \cos x)^2} = \large\frac{2 \sin x}{\bigg( 1+\Large\frac{1}{\sec x} \bigg)^2}$$ =\large\frac{2 \sin x}{ \Large\frac{(\sec x+1)^2}{\sec^2 x} }$
$= \large\frac{2 \sin x \sec^2x}{(\sec x+1)^2}$
$= \large\frac{ \Large\frac{2\sin x}{\cos x}\sec x}{(\sec x+1)^2}$
$= \large\frac{2 \sec x \tan x }{( \sec x+1)^2}$
edited Apr 14, 2014