Question

please solve with solution ​

Answer 1

Given,

$\sf \: \dfrac{d}{dx} \bigg \{ \dfrac{a - b \cos(x) }{a + b \cos(x) } \bigg \}$

Consider

$\sf \: y = \dfrac{a - b \cos(x) }{a + b \cos(x) }$

Taking log on both sides,

$\implies \sf \: log(y) = log \bigg| \dfrac{a - b \cos(x) }{a + b \cos(x) } \bigg | \\ \\ \implies \sf \: log(y) = log |a - b \cos(x) | - log |a + b \cos(x) |$

Differentiating w.r.t x (Chain Rule) -

$\implies \sf \: \dfrac{1}{y} \times \dfrac{dy}{dx} = \dfrac{b \sin(x) }{a - b \cos(x) } - \dfrac{ - b \sin(x) }{a + b \cos(x) } \\ \\ \implies \sf \: \dfrac{1}{y} \times \dfrac{dy}{dx} = \dfrac{b \sin(x) (a + b \cos(x)) + b \sin(x) ( a - b \cos(x) ) }{(a + b \cos(x))(a - b \cos(x) ) } \\ \\ \implies \sf \: \dfrac{dy}{dx} = \bigg \{ \dfrac{2ab \sin(x) }{(a + b \cos(x))(a - b \cos(x) )} \bigg \} \times \dfrac{a - b \cos(x) }{a + b \cos(x) } \\ \\ \implies \boxed{ \boxed{\sf \: \dfrac{dy}{dx} = \dfrac{2ab \sin(x) }{(a + b \cos(x)) {}^{2} } }}$

Option (b) is correct.