Question

Differentiate sin^-1 ( (a + b cosx) / (b + a cosx) ) , b>a w.r.t. x.

Answer 1

 Given, y = sin^-1 (( a + b cos x ) / ( b + a cos x ) )


          sin y = ( a + b cos x ) / ( b + a cos x ) 


          Differentiating on both sides, we get


cos y * dy / dx = ( ( b + a cos x ) * ( - b sin x ) - ( a + b cos x ) ( - a sin x ) ] / ( b + a  cos x )^2 


cos y * dy / dx = ( -b ^ 2 sin x - ab sin x cos x + a ^ 2 sin x + ab sin x cos x ) / ( b + a cos x )^2 


cos y * dy / dx = ( a ^ 2 - b ^ 2 ) sin x / ( b + a cos x )^2 


dy / dx = ( a ^ 2 - b ^ 2 ) ( sin x / ( b + acos x ) ^ 2 ) * ( 1 / sqrt ( 1 - sin ^ 2y ) ) 


dy / dx = ( a ^ 2 - b ^ 2 ) * sin x / ( b + acos x )^2 * ( ( b + acos x ) / ( sin x sqrt ( b^2 - a^2 ) ) )


dy / dx = - sqrt ( b^2 - a^2 ) / ( b + a cos x ).


Hope this helps!