Calculus problem #2661

Applying the quotient rule $f'({x})=\frac{120sinx+120sin^2x+120cos^2x}{(1+sinx)^2}$

Which simplifies to $f'({x})=\frac{120}{1+sinx}$

Thus $f'({\frac{\pi}{6}})=\frac{120}{1+sin(\frac{\pi}{6})}$

and $f'({\frac{\pi}{6}})=\frac{240}{3}$

Therefore $f'({\frac{\pi}{6}})=\boxed{80}$

When we apply the quotient rule, we get f'(x)={(1+sinx)120sinx+120cosx(cosx)}/(1+sinx)^2 On putting x=π/6 we get f'(π/6)={(90+90)×4}/9=80