Question

y=√a^2cos^x+b^2sin^x then dy/dx=?​

Answer 1

Step-by-step explanation:

y=√(a^2 cos^2⁡〖x +〗 b^2 sin^2⁡〖x 〗 )

y^2= a^2 cos^2⁡〖x+ b^2 sin^2 x〗 ,

or 2y dy/dx=-2a^2 cosx .sinx+2b^2 sinx.cosx=(b^2-a^2 )2sinx.cosx=(b^2-a^2 )sin2x

dy/dx={(b^2-a^2 )sin2x}/2y={(b^2-a^2 )sin2x }/(2√(a^2 〖cos^2⁡〖x+b^2 sin 〗〗^2 x))