Answer to Question #313454 in Calculus for justt

Question

Find the Derivative of the following function

1. y = x² cos x - 2xsinx - 2cosx
2. x(cos2x + 3y) = ysinx

Answer 1

$1.\\ y=x^2cosx-2xsinx-2cosx\\ y'=2xcosx+x^2(-sinx)-2sinx-\\-2xcosx-2(-sinx)=-x^2sinx\\ 2.\\ x(cos2x+3y)=ysinx\\ y=\frac{xcos2x}{sinx-3x}\\ y'=\frac{(cos2x+x(-2sin2x))(sinx-3x)-xcos2x(cosx-3)}{(sinx-3x)^2}=\\ =\frac{(cos2x-2xsin2x)(sinx-3x)-xcos2x(cosx-3)}{(sinx-3x)^2}$