What is the derivative of sin(x/pi)?

Mar 8, 2018

$y = \sin \left(\frac{x}{\pi}\right) \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(\frac{x}{\pi}\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{x}{\pi}\right) = \frac{1}{\pi} \cdot \cos \left(\frac{x}{\pi}\right)$

Explanation:

We have,
*color(blue)(sinC-sinD=2cos((C+D)/2)sin((C-D)/2)
$\textcolor{red}{{f}^{'} \left(x\right) = {\lim}_{t \to x} \frac{f \left(t\right) - f \left(x\right)}{t - x}}$
Here, $f \left(x\right) = \sin \left(\frac{x}{\pi}\right)$
So,
${f}^{'} \left(x\right) = {\lim}_{t \to x} \frac{\sin \left(\frac{t}{\pi}\right) - \sin \left(\frac{x}{\pi}\right)}{t - x}$
$= {\lim}_{t \to x} \frac{2 \cos \left(\frac{\frac{t}{\pi} + \frac{x}{\pi}}{2}\right) \sin \left(\frac{\frac{t}{\pi} - \frac{x}{\pi}}{2}\right)}{t - x}$
$= {\lim}_{t \to x} 2 \cos \left(\frac{t + x}{2 \pi}\right) {\lim}_{t \to x} \left(\sin \frac{\frac{t - x}{2 \pi}}{t - x}\right)$
=2cos((x+x)/(2pi))lim_((t-x) to 0)(sin((t-x)/(2pi))/(((t-x)/(2pi)))*(1/(2pi))
$= \cancel{2} \cos \left(\frac{x}{\pi}\right) \left(1\right) \left(\frac{1}{\cancel{2} \pi}\right) = \frac{1}{\pi} \cdot \cos \left(\frac{x}{\pi}\right)$