## University Calculus: Early Transcendentals (3rd Edition)

$\frac{ds}{du}=\frac{9}{2}$
Given $s=t^2+5t$ and $t=({u^2+2u})^{\frac{1}{3}}$ On differentiating both sides: $\frac{ds}{du}=(2t+5)\frac{dt}{du}$ $\frac{dt}{du}=\frac{1}{3}(({u^2+2u})^{\frac{-2}{3}}\frac{d({u^2+2u})}{du})$ $\frac{dt}{du}=\frac{(2u+2)}{3}(({u^2+2u})^{\frac{-2}{3}}$ on putting the above value in $\frac{ds}{du}$ $\frac{ds}{du}=(2({u^2+2u})^{\frac{1}{3}}+5)\frac{(2u+2)}{3}(({u^2+2u})^{\frac{-2}{3}}$ at $u=2$ $\frac{ds}{du}=(2({2^2+2(2)})^{\frac{1}{3}}+5)\frac{(2(2)+2)}{3}(({(2)^2+2(2)})^{\frac{-2}{3}}$ $\frac{ds}{du}=(2({8})^{\frac{1}{3}}+5)\frac{(6)}{3}(({8})^{\frac{-2}{3}}$ $\frac{ds}{du}=\frac{9}{2}$