Find y′y'y′ for y=xsinx1+x2y = \frac{x\sin x}{1 + x^2}y=1+x2xsinx.
(sinx+xcosx)(1+x2)−2x2sinx(1+x2)2\frac{(\sin x + x\cos x)(1+x^2) - 2x^2\sin x}{(1+x^2)^2}(1+x2)2(sinx+xcosx)(1+x2)−2x2sinx
sinx+xcosx1+x2\frac{\sin x + x\cos x}{1+x^2}1+x2sinx+xcosx
(sinx+xcosx)(1+x2)−xsinx⋅2x(1+x2)2\frac{(\sin x + x\cos x)(1+x^2) - x\sin x \cdot 2x}{(1+x^2)^2}(1+x2)2(sinx+xcosx)(1+x2)−xsinx⋅2x
Both (a) and (c) are the same