Question

If f(x) = (sin x)*, find f'(pi/2).

Answer 1

$f(x) = sin^x(x)$

We want f'(x)

So, lets derive, we can consider sin(x) as an normal x and derive it as x² going to 2x, and apply chain rule to sin(x)

$f'(x)=x.sin^{x-1}(x).cos^x(x)$

So just use x as pi/2

f'(π/2) = π/2 . sin^(π/2-1)(π/2) . cos^(π/2).(π/2)

sin(π/2) = 1

cos(π/2) = 0

f'(π/2) = π/2 . 1^(π/2 - 1) . 0^(π/2)

f'(π/2) = 0