The problem statement

Find the derivative $y_{x}^{\prime}$.

$$
\left\{\begin{array}{l}
x=\sqrt{1-t^{2}} \\
y=\tan \sqrt{1+t}
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=\left(\sqrt{1-t^{2}}\right)^{\prime}=\frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)=-\frac{t}{\sqrt{1-t^{2}}}$

$y_{t}^{\prime}=(\tan \sqrt{1+t})^{\prime}=\frac{1}{\cos ^{2}(\sqrt{1+t})} \cdot \frac{1}{2 \sqrt{1+t}}=\frac{1}{2 \cos ^{2}(\sqrt{1+t}) \sqrt{1+t}}$

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{1}{2 \cos ^{2}(\sqrt{1+t}) \sqrt{1+t}}\right) /\left(-\frac{t}{\sqrt{1-t^{2}}}\right)=\frac{\sqrt{1-t^{2}}}{-2 t \cdot \cos ^{2}(\sqrt{1+t}) \sqrt{1+t}}= \\
& =-\frac{\sqrt{1-t}}{2 t \cdot \cos ^{2}(\sqrt{1+t})}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 16-2