Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 48

Find the first and second derivatives of $y = \sin^2(\pi t)$
Solving for the first derivative of the given function


$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dt} \left[ \sin^2 (\pi t) \right]\\
\\
y' &= \frac{d}{dt} \left[ \sin (\pi t) \right]^2\\
\\
y' &= 2[\sin(\pi t)] \cdot \frac{d}{dt} [\sin(\pi t)]\\
\\
y' &= 2 \sin (\pi t) \cos (\pi t) \cdot \frac{d}{dt} (\pi t) \\
\\
y' &= 2 \sin (\pi t) \cos (\pi t) (\pi) && \left( \text{Recall the Double Angle Formula } (\sin 2x = 2\sin x \cos x)\right)\\
\\
y' &= \pi \sin ( 2 \pi t)
\end{aligned}
\end{equation}
$


Solving for the second derivative of the given function

$
\begin{equation}
\begin{aligned}
y'' &= \frac{d}{dt} \left[ \pi \sin(2 \pi t) \right]\\
\\
y'' &= (\pi) \frac{d}{dt} [\sin (2\pi t)]\\
\\
y'' &= (\pi) \left[ \cos (2\pi t) \cdot \frac{d}{dt} (2\pi t) \right]\\
\\
y'' &= (\pi) \left[ \cos (2\pi t) \cdot 2\pi \frac{d}{dt}(t)\right]\\
\\
y'' &= (\pi) \left[ \cos (2\pi t)(2\pi) (1) \right]\\
\\
y'' &= (\pi) \left[ 2 \pi \cos(2 \pi t) \right]\\
\\
y'' &= 2 \pi^2 \cos ( 2 \pi t)
\end{aligned}
\end{equation}
$