## Task Condition

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

$$
\left\{\begin{array}{l}
x=\cos t \\
y=\sin ^{4}\left(\frac{t}{2}\right)
\end{array}\right.
$$
## Solution

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

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

We obtain:

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

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

$=-\frac{2 \sin (t / 2) \cdot \cos ^{2}(t / 2)+\sin ^{3}(t / 2)}{2 \cos ^{2}(t / 2)}$

Then:

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

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

## Problem Kuznetsov Differentiation 20-24