Answer
$\frac{t}{\sqrt (t^{2}+9)}$
Work Step by Step
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,
$\frac{dy}{dt}=\frac{dy}{ds}.\frac{ds}{dt}$= $\frac{1}{2√(t^{2}+9)}2t$= $\frac{t}{\sqrt (t^{2}+9)}$
