Answer
\[(a)\; \frac{dy}{dt}=1\]
\[(b)\; \frac{dx}{dt}=25\]
Work Step by Step
$y=\sqrt{2x+1}$ ____(1)
Since $x$ and $y$ are functions of $t$
Differentiating (1) with respect to $t$
$\large\frac{dy}{dt}=\frac{2}{2\sqrt{2x+1}}\frac{dx}{dt}$
$\large\frac{dy}{dt}=\frac{1}{\sqrt{2x+1}}\frac{dx}{dt}$ ___(2)
$(a)\;\;$ $\large\frac{dx}{dt}$=$3, \;\;x=4$
From (2)
$\large\frac{dy}{dt}=\frac{1}{\sqrt{2(4)+1}}\cdot$ $(3)$
$\large\frac{dy}{dt}=\frac{1}{3}\cdot$ $ 3=1$
Hence, $ \large\frac{dy}{dt}=1$
$(b)\;\;$ $\large \frac{dy}{dt}=5\;\;$ ,$x=12$
$5=\large\frac{1}{\sqrt{2(12)+1}} \frac{dx}{dt}$
$\large \frac{dx}{dt}=$ $5\times 5=25$
Hence, $ \large\frac{dx}{dt}=$ $25$.
