Question

Q. if y = sin^(-1){ (a + b cosx) / ( b + a cosx)} then find dy/dx ​

Answer 1

Step-by-step explanation:

We have,

$y = \sin^{ - 1} \bigg \{ \frac{a + b \cos(x) }{b + a \cos(x) } \bigg \}$

$\implies \frac{dy}{dx} = \frac{1}{ \sqrt{1 - ( \frac{a + b \cos(x) }{b + a \cos(x) } )^{2}} }. \frac{d}{dx} \bigg \{ \frac{a + b \cos(x) }{b + a \cos(x) } \bigg \}$

$\implies \frac{dy}{dx} = \frac{b + a \cos(x) }{ \sqrt{(b + a \cos(x))^{2} - ( a + b \cos(x) )^{2}} }. \bigg \{ \frac{(b + a \cos(x) ).\frac{d}{dx}( a + b \cos(x) ) - (a + b \cos(x) ). \frac{d}{dx}(b + a \cos(x) )}{(b + a \cos(x) )^{2} } \bigg \}$

$\implies \frac{dy}{dx} = \frac{1 }{ \sqrt{(b) ^{2} + (a \cos(x))^{2} - ( a) ^{2} - ( b \cos(x) )^{2}} }. \frac{ \{- b \sin(x) (b + a \cos(x) ) + (a \sin(x) ) (a + b \cos(x) ) \}}{(b + a \cos(x) ) }$

$\implies \frac{dy}{dx} = \frac{1 }{ \sqrt{b^{2} + a^{2} \cos^{2} (x) - a ^{2} - b^{2} \cos^{2} (x) } }. \frac{ \{- b ^{2} \sin(x) - ab \sin(x) \cos(x) + a ^{2} \sin(x) + ab \sin(x) \cos(x)\}}{(b + a \cos(x) ) }$

$\implies \frac{dy}{dx} = \frac{1 }{ \sqrt{b^{2} \sin^{2} (x) - a^{2} \sin^{2} (x) } }. \frac{ \{- b ^{2} \sin(x) + a ^{2} \sin(x) \}}{(b + a \cos(x) ) }$

$\implies \frac{dy}{dx} = \frac{- \{ b ^{2} \sin(x) - a ^{2} \sin(x) \}}{ (b + a \cos(x) ) \sqrt{b^{2} \sin^{2} (x) - a^{2} \sin^{2} (x) } }$