Condition of the problem

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

$$
\left\{\begin{array}{l}
x=\sin t \\
y=\sec t
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=(\sin t)^{\prime}=\cos t$

$y_{t}^{\prime}=(\sec t)^{\prime}=\left(\frac{1}{\cos t}\right)^{\prime}=-\frac{1}{\cos ^{2} t} \cdot(-\sin t)=\frac{\sin t}{\cos ^{2} t}$

We obtain:

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

$\left(y_{x}^{\prime}\right)_{t}^{\prime}=\left(\frac{\sin t}{\cos ^{3} t}\right)^{\prime}=\frac{\cos t \cdot \cos ^{3} t-\sin t \cdot 3 \cos ^{2} t \cdot(-\sin t)}{\cos ^{6} t}=$

$=\frac{\cos ^{2} t+3 \sin ^{2} t}{\cos ^{4} t}=\frac{1+2 \sin ^{2} t}{\cos ^{4} t}$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{1+2 \sin ^{2} t}{\cos ^{4} t}\right) / \cos t=\frac{1+2 \sin ^{2} t}{\cos ^{5} t}$