Suppose f(𝜋/3) = 4 and f '(𝜋/3) = −3, and let g(x) = f(x) sin(x) and h(x) = cos(x)/f(x). Find g'(𝜋/3) Find h'(𝜋/3)

Question

Don't know how to do my calc homework pls help.

Answer 1

Notice that g(x) is a product (it's f(x) times sin(x)) and notice h(x) is a quotient (it's cos(x) divided by f(x)) so you can use the product rule for g(x) and the quotient rule for h(x):

Product rule: for u•v, (u•v)' = u'v + uv' where u = f(x) and v = sin(x). Since we are looking for g'(π/3) then:

g'(π/3) = f'(π/3)•sin(π/3) + f(π/3)•[sin(π/3)]'

we are told that f '(π/3) = -3, by the unit circle we know that sin(π/3) = √3/2.

we are also told that f(π/3) = 4 and since the derivative of sin(x) = cos(x) then [sin(π/3)]' = cos(π/3) which is 1/2 so:

g'(π/3) = (-3)•√3/2 + (4)(1/2)

g'(π/3) = -3√3/2 + 2

You do the same with the quotient rule:

for u/v, (u/v)' = (u'v - uv')/v²

Answer 2

For g'(x) use the Product Rule.

For h'(x) use the Quotient Rule