Question

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed nonzero constants and m and n are integers): $\frac{sec\,x-1}{sec\,x+1}$

Answer 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'(x) = (1+cosx) $\frac{d}{dx}$(1-cosx) - (1-cosx) $\frac{d}{dx}$ $\frac{1-cos\,x}{(1+cos\,x)^2}$
=$\frac{(1+cos\,x)(sin\,x)-(1-cos\,x)(-sin\,x)}{(1+cos\,x)^2}$
= $\frac{2sin\,x}{(1+cos\,x)^2}$
=$\frac{2sin\,x}{(1+\frac{1}{sec\,x})^2}$ = $\frac{2sin\,x}{\frac{(sec\,x+1)^2}{sec^2x}}$
= $\frac{2sin\,xsec^2x}{(sec\,x+1)^2}$
=$\frac{2sec\,x\,tan\,x}{(sec\,x+1)^2}$