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

To find h'(x), differentiate the function h(x) = (cos x)/f(x) using the product rule.

Step-by-step explanation:

To find h'(x), we need to differentiate the function h(x) = (cos x)/f(x).

First, let's find the derivative of cos x, which is -sin x.

Next, we need to find the derivative of f(x). Since f(π/3) = 3 and f '(π/3) = −7, we know the slope of the tangent line at x = π/3 is -7.

Using the product rule, we can now differentiate h(x) = (cos x)/f(x) as follows:

h'(x) = [f(x)(-sin x) - cos x(f '(x))]/[f(x)]^2

Answer 2

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.