## Calculus 10th Edition

The derivative is$\frac{(1-5x^3)(\sqrt x)}{2x(x^3+1)^2}$.
Using the quotient rule: $h'(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=\sqrt x; u'(x)=\frac{1}{2\sqrt x}$ $v(x)=x^3+1; v'(x)=3x^2$ $h'(x)=\frac{(\frac{1}{2\sqrt x})(x^3+1)-(\sqrt x)(3x^2)}{(x^3+1)^2}=$ $\frac{(\frac{1}{2\sqrt x})(x^3+1)-(\frac{2x}{2\sqrt x})(3x^2)}{(x^3+1)^2}=$ $\frac{x^3+1-6x^3}{(2\sqrt x)((x^3+1)^2)}=$ $\frac{(1-5x^3)(\sqrt x)}{2x(x^3+1)^2}$