Suppose f (π/3) = 3 and f ' (π/3) = - 7,

and let

g (x) = f (x) sin x

and

h (x) = (cos x) / f (x).

Find the h' (x)

Et's give this a go:h (x) = cos (x) / f (x)

derivative (recall the quotient rule) h' (x) = [ f (x) * (-sin (x)) - cos (x) * f' (x) ] / [ f (x) ]^2

simplifyh' (x) = [ - sin (x) * f (x) - cox (x) * f ' (x) ] / [ f (x) ]^2h' (π/3) = [ - sin (π/3) * f (π/3) - cox (π/3) * f ' (π/3) ] / [ f (π/3) ]^2h' (π/3) = - (3-√/2) ∗ (3) - (1/2) ∗ (-7) / (3) 2

h' (π/3) = (-33-√/2+7/2) / 9

And you can further simplify if you want, I'll stop there.