If `y=asinx+bcosx`, then `y^2+((dy)/(dx))^2` is a

Options

A. function of x
B. function of y
C. function x and y
D. constant

Answer

D

Explanation

`y=asinx+bcosx`
Differentiating with respect to x, we get
`(dy)/(dx)=acosx-bsinx`
`Now((dy)/(dx))^(2)=(acosx-bsinx)^(2)`
`=a^(2)cos^(2)x+b^(2)sin^(2)x-2absinxcosx`
and `y^(2)=(asinx+bcosx)^(2)`
`=a^(2)sin^(2)x+b^(2)cos^(2)x+2absinxcosx`
So, `((dy)/(dx))^(2)+y^(2)=a^(2)(sin^(2)x+cos^(2)x)+b^(2)(sin^(2)x+cos^(2)x)`
Hence `((dy)/(dx))^(2)+y^(2)=(a^(2)+b^(2))=` constant