Answer
$\frac{ds}{du}=\frac{9}{2}$
Work Step by Step
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 $u=2$
$\frac{ds}{du}=(2({2^2+2(2)})^{\frac{1}{3}}+5)\frac{(2(2)+2)}{3}(({(2)^2+2(2)})^{\frac{-2}{3}}$
$\frac{ds}{du}=(2({8})^{\frac{1}{3}}+5)\frac{(6)}{3}(({8})^{\frac{-2}{3}}$
$\frac{ds}{du}=\frac{9}{2}$
