Answer
$$
\sqrt {u}+\sqrt {2v+1}=5;
$$
(a)
$$
\begin{aligned}
\frac{d u}{d v} &=-\frac{2 u^{1 / 2}}{(2 v+1)^{1 / 2}} .
\end{aligned}
$$
(b)
$$
\begin{aligned}
\frac{d v}{d u} &=-\frac{(2 v+1)^{1 / 2}}{2 u^{1 / 2}} .
\end{aligned}
$$
Work Step by Step
$$
\sqrt {u}+\sqrt {2v+1}=5;
$$
(a) $\frac{d u}{d v}$
Now, we can calculate $\frac{d u}{d v}$ by implicit differentiation,
$$
\begin{aligned}
\frac{d }{d v}(\sqrt{u}+\sqrt{2 v+1}) &=\frac{d }{d v}(5) \\ \frac{1}{2} u^{-1 / 2} \frac{d u}{d v}+\frac{1}{2}(2 v+1)^{-1 / 2}(2) &=0 \\
\frac{1}{2} u^{-1 / 2} \frac{d u}{d v} &=-\frac{1}{(2 v+1)^{1 / 2}} \\ \frac{d u}{d v} &=-\frac{2 u^{1 / 2}}{(2 v+1)^{1 / 2}} .
\end{aligned}
$$
$$
\sqrt {u}+\sqrt {2v+1}=5;
$$
(b) $\frac{d v}{d u}$
$$
\begin{aligned}
\frac{d }{d u}(\sqrt{u}+\sqrt{2 v+1}) &=\frac{d }{d u}(5) \\ \frac{1}{2} u^{-1 / 2}+\frac{1}{2}(2 v+1)^{-1 / 2}(2) \frac{d v}{d u} &=0 \\(2 v+1)^{-1 / 2} \frac{d v}{d u} &=-\frac{1}{2} u^{-1 / 2} \\ \frac{d v}{d u} &=-\frac{(2 v+1)^{1 / 2}}{2 u^{1 / 2}} .
\end{aligned}
$$
