Calculus: Early Transcendentals 8th Edition

$$\frac{dg}{du}=\frac{48(u^3-1)^7u^2}{(u^3+1)^9}$$
$$g(u)=\Bigg(\frac{u^3-1}{u^3+1}\Bigg)^8$$ Then $$\frac{dg}{du}=\frac{d(\frac{u^3-1}{u^3+1})^8}{du}$$ Let $x=\frac{u^3-1}{u^3+1}$, then $g(x)=x^8$. According to the Chain Rule, $$\frac{dg}{du}=\frac{dg}{dx}\frac{dx}{du}$$ $$\frac{dg}{du}=\frac{d(x^8)}{dx}\frac{d(\frac{u^3-1}{u^3+1})}{du}$$ *Find $\frac{d(x^8)}{dx}$ $$\frac{d(x^8)}{dx}=8x^7$$ * Find $\frac{d(\frac{u^3-1}{u^3+1})}{du}$ $$\frac{d(\frac{u^3-1}{u^3+1})}{du}=\Bigg(\frac{u^3-1}{u^3+1}\Bigg)'$$ $$\frac{d(\frac{u^3-1}{u^3+1})}{du}=\frac{(u^3-1)'(u^3+1)-(u^3-1)(u^3+1)'}{(u^3+1)^2}$$ $$\frac{d(\frac{u^3-1}{u^3+1})}{du}=\frac{3u^2(u^3+1)-3u^2(u^3-1)}{(u^3+1)^2}$$ $$\frac{d(\frac{u^3-1}{u^3+1})}{du}=\frac{3u^2\times2}{(u^3+1)^2}$$ $$\frac{d(\frac{u^3-1}{u^3+1})}{du}=\frac{6u^2}{(u^3+1)^2}$$ Therefore, $$\frac{dg}{du}=8x^7\times\frac{6u^2}{(u^3+1)^2}$$ $$\frac{dg}{du}=8\Bigg(\frac{u^3-1}{u^3+1}\Bigg)^7\times\frac{6u^2}{(u^3+1)^2}$$ $$\frac{dg}{du}=\frac{48(u^3-1)^7u^2}{(u^3+1)^9}$$