Differentiate wrt. x

sin^3x+cos^6x

Let y = sin^3x+cos^6x

= u+v [ u = sin^3x ; v = cos^3x ]


dy/dx= du/dx+dv/dx


now, du/dx = d/dx(sin^3x) = 3sin^2x.cosx ..............(1)


again, dv/dx= d/dx(cos^6) = 6cos^5(-sinx)...............(2)


(1)+(2) = 3sin^2cosx-6cos^5xsinx

= 3sinxcosx(sinx-2cos^4x)

therefore, dy/dx = 3sinx.cosx.(sin.x - 2cos^4x) Ans.

let y = sin^3x+cos^6x
let u = sin^3x
v = cos^6x
ie, y=u+v

dy/dx= du/dx+dv/dx

du/dx = d/dx(sin^3x) = 3sin^2x.cosx ..............(1)

dv/dx= d/dx(cos^6x) = 6cos^5x.(-sinx)...............(2)


(1)+(2) = 3sin^2cosx-6cos^5xsinx

therefore, dy/dx= 3.sinx.cosx(sinx-2.cos^4x)

3sin^2 x (cos x) - 6cos ^5 x sin x
3sin x cos x (sin x cos ^ 4 x)