## Problem Statement

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

$$
\left\{\begin{array}{l}
x=\left(1+\cos ^{2} t\right)^{2} \\
y=\frac{\cos t}{\sin ^{2} t}
\end{array}\right.
$$
## Solution

$$
\begin{aligned}
& x_{t}^{\prime}=\left(\left(1+\cos ^{2} t\right)^{2}\right)^{\prime}=2\left(1+\cos ^{2} t\right) \cdot 2 \cos t \cdot(-\sin t)= \\
& =-4\left(1+\cos ^{2} t\right) \cdot \cos t \cdot \sin t \\
& y_{t}^{\prime}=\left(\frac{\cos t}{\sin ^{2} t}\right)^{\prime}=\frac{-\sin t \cdot \sin ^{2} t-\cos t \cdot 2 \sin t \cdot \cos t}{\sin ^{4} t}= \\
& =\frac{-\sin t \cdot\left(\sin ^{2} t+2 \cos ^{2} t\right)}{\sin ^{4} t}=-\frac{\sin ^{2} t+2 \cos ^{2} t}{\sin ^{3} t}
\end{aligned}
$$

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{\sin ^{2} t+2 \cos ^{2} t}{\sin ^{3} t}\right) /\left(-4\left(1+\cos ^{2} t\right) \cdot \cos t \cdot \sin t\right)= \\
& =\frac{\sin ^{2} t+2 \cos ^{2} t}{\sin ^{3} t \cdot 4\left(1+\cos ^{2} t\right) \cdot \cos t \cdot \sin t}=\frac{1+\cos ^{2} t}{\sin ^{3} t \cdot 4\left(1+\cos ^{2} t\right) \cdot \cos t \cdot \sin t}= \\
& =\frac{1}{4 \sin ^{4} t \cdot \cos t}
\end{aligned}
$$

Kuznetsov Differentiation 16-15