Answer
(a) The derivative of
$$
y=f(t)=\tan \sqrt{t}
$$
is
$$
f^{\prime}(t)=\sec ^{2} \sqrt{t} \cdot \frac{1}{2} t^{-1 / 2}=\frac{\sec ^{2} \sqrt{t}}{2 \sqrt{t}},
$$
so
$$
d y=\frac{\sec ^{2} \sqrt{t}}{2 \sqrt{t}} d t
$$
(b) The derivative of
$$
y=f(v)=\frac{1-v^{2}}{1+v^{2}}
$$
is
$$
\begin{aligned}
f^{\prime}(v)&=\frac{\left(1+v^{2}\right)(-2 v)-\left(1-v^{2}\right)(2 v)}{\left(1+v^{2}\right)^{2}} , \text{[ the Quotient Rule]}\\
&=\frac{-2 v\left[\left(1+v^{2}\right)+\left(1-v^{2}\right)\right]}{\left(1+v^{2}\right)^{2}} \\
&=\frac{-2 v(2)}{\left(1+v^{2}\right)^{2}}\\
&=\frac{-4 v}{\left(1+v^{2}\right)^{2}},
\end{aligned}
$$
so
$$
d y=\frac{-4 v}{\left(1+v^{2}\right)^{2}} d v
$$
Work Step by Step
(a) The derivative of
$$
y=f(t)=\tan \sqrt{t}
$$
is
$$
f^{\prime}(t)=\sec ^{2} \sqrt{t} \cdot \frac{1}{2} t^{-1 / 2}=\frac{\sec ^{2} \sqrt{t}}{2 \sqrt{t}},
$$
so
$$
d y=\frac{\sec ^{2} \sqrt{t}}{2 \sqrt{t}} d t
$$
(b) The derivative of
$$
y=f(v)=\frac{1-v^{2}}{1+v^{2}}
$$
is
$$
\begin{aligned}
f^{\prime}(v)&=\frac{\left(1+v^{2}\right)(-2 v)-\left(1-v^{2}\right)(2 v)}{\left(1+v^{2}\right)^{2}} , \text{[ the Quotient Rule]}\\
&=\frac{-2 v\left[\left(1+v^{2}\right)+\left(1-v^{2}\right)\right]}{\left(1+v^{2}\right)^{2}} \\
&=\frac{-2 v(2)}{\left(1+v^{2}\right)^{2}}\\
&=\frac{-4 v}{\left(1+v^{2}\right)^{2}},
\end{aligned}
$$
so
$$
d y=\frac{-4 v}{\left(1+v^{2}\right)^{2}} d v
$$
