Calculus (3rd Edition)

$\frac{t}{\sqrt (t^{2}+9)}$
Given y= $(t^{2}+9)^{1/2}$ Put $t^{2}+9$= s Then, $\frac{ds}{dt}= 2t$ As y= $s^{1/2}$, we have $\frac{dy}{ds}= \frac{1}{2}×s^{-\frac{1}{2}}= \frac{1}{2√s}$ Substituting the value of s, we get $\frac{dy}{ds}= \frac{1}{2√(t^{2}+9)}$ According to the chain rule, $\frac{dy}{dt}=\frac{dy}{ds}.\frac{ds}{dt}$= $\frac{1}{2√(t^{2}+9)}2t$= $\frac{t}{\sqrt (t^{2}+9)}$