Question

Differentiate `(a^2sin^2x+b^2cos^2x)`?

Answer 1

`d/(dx) (a^2sin^2x+b^2cos^2x)=sin2x(a^2-b^2)`

Explanation:

`d/(dx) (a^2sin^2x+b^2cos^2x)`

= `d/(dx) a^2sin^2x+d/(dx) b^2cos^2x`

= `a^2xx2sinx xx cosx+b^2xx2cosx xx(-sinx)`

= `2a^2sinxcosx-2b^2sinxcosx`

= `sin2x(a^2-b^2)`