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