Question

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

Answer 1

`y=sin(x/pi)=>(dy)/(dx)=cos(x/pi)*d/(dx)(x/pi)=1/pi*cos(x/pi)`

Explanation:

We have,
`*color(blue)(sinC-sinD=2cos((C+D)/2)sin((C-D)/2)`
`color(red)(f^'(x)=lim_(t to x)(f(t)-f(x))/(t-x))`
Here, `f(x)=sin(x/pi)`
So,
`f^'(x)=lim_(t to x)(sin(t/pi)-sin(x/pi))/(t-x)`
`=lim_(t to x)(2cos((t/pi+x/pi)/2)sin((t/pi-x/pi)/2))/(t-x)`
`=lim_(t to x)2cos((t+x)/(2pi))lim_(t to x)(sin((t-x)/(2pi))/(t-x))`
`=2cos((x+x)/(2pi))lim_((t-x) to 0)(sin((t-x)/(2pi))/(((t-x)/(2pi)))*(1/(2pi))`
`=cancel(2)cos(x/pi)(1)(1/(cancel(2)pi))=1/pi*cos(x/pi)`