## Task Condition

Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.

$$
\left\{\begin{array}{l}
x=\cos ^{2} t \\
y=\operatorname{tg}^{2} t
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=\left(\cos ^{2} t\right)^{\prime}=2 \cos t \cdot(-\sin t)=-2 \cos t \cdot \sin t$

$y_{t}^{\prime}=\left(\operatorname{tg}^{2} t\right)^{\prime}=2 \operatorname{tg} t \cdot \frac{1}{\cos ^{2} t}=\frac{2 \sin t}{\cos ^{3} t}$

We obtain:

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

$\left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(-\frac{1}{\cos ^{4} t}\right)^{\prime}=-\frac{-4}{\cos ^{5} t} \cdot(-\sin t)=-\frac{4 \sin t}{\cos ^{5} t}$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{4 \sin t}{\cos ^{5} t}\right) /(-2 \cos t \cdot \sin t)=\frac{2}{\cos ^{6} t}$

Kuznetsov Differentiation Problem 20-16