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Answer 1

$\text{Let f(x) = \frac{secx-1}{secx+1} }\\\\\\\text{Let u = secx - 1 and v = secx + 1 }\\\\\\\text{ \therefore f(x) = \frac{u}{v} }\\\\\\\text{So, f'(x) = (\frac{u}{v})' }\\\\\\\text{Using quotient rule :}\\\\\\\text{f'(x) = \frac{u'v - v'u}{v^2} }\\\\\\\text{Finding u' and v' : }\\\\\\\text{u = secx - 1 }\\\\\\ \text{u' = (secx - 1)'}\\\\\\ \text{= secx tanx - 0}\\\\\\ \text{=secx tanx}\\\\\\\text{and , v = secx + 1 }\\\\\\ \text{v' = secx tanx + 0 }\\\\\\ \text{= secx tanx}$

$\text{Now, f'(x) = (\frac{u}{v})' }\\\\\\ \text{= \frac{u'v - v'u}{v^2} }\\\\\\=\frac{(secx tanx)(secx+1)-(secx tanx)(secx-1)}{(secx+1)^2} \\\\\\= \frac{secx.tanx[(secx+1)-(secx-1)]}{(secx+1)^2} \\\\\\= \frac{secx.tanx(secx+1-secx+1)}{(secx+1)^2} \\\\\\= \frac{secx.tanx(2)}{(secx+1)^2}\\\\\\ =\boxed{ \frac{2secx.tanx}{(secx+1)^2}}$

Answer 2

Answer:

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