Question

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)

Answer 1

i know its the quotient rule
this is what i got: -7cos(x)+sin(x)-3/(cosx^2)

Answer 2

but thats incorrect

Answer 3

you're sure it's not find h'(π/3) ?

Answer 4

yeah its h'(pi/3)
sorry

Answer 5

why is it incorrect? did i make a simple error?

Answer 6

Let'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

simplify
h'(x) = [ -sin(x)*f(x) - cox(x)*f '(x) ] / [ f(x) ]^2
h'(π/3) = [ -sin(π/3)*f(π/3) - cox(π/3)*f '(π/3) ] / [ f(π/3) ]^2
h'(π/3) = \[ -(\sqrt{3}/2)*(3) -(1/2)*(-7) /(3)^2\]
h'(π/3) = \[(-3\sqrt{3}/2+7/2)/9\]

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

Answer 7

thank you