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 the following.

Question

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 the following

g'(𝜋/3)

h'(𝜋/3)

Answer 1

Hi, Julia!

The keys to solving this problem are the product and quotient rules for differentiation.

Let's start with the product rule; say you have two functions of x, a(x) and b(x). The following equation gives us the product rule: (ab)' = a'b+ab'

We can use this rule to solve for g'(x)

Let a(x) = f(x), b(x) = sin(x). --> We can take the derivatives of a and b individually, which yields a'(x) = f'(x) and b'(x) = cos(x).

Note here that, based on these choices for a(x) and b(x), we have g(x) = f(x)sin(x) = a(x)b(x)

Applying the product rules yields g'(x) = a'(x)b(x) + a(x)b'(x) --> g'(x) = f'(x)sin(x) + f(x)cos(x)

Now that we have a formula for g'(x), all we need to do is plug in x=π/3

g'(π/3) = f'(π/3)sin(π/3) + f(π/3)cos(π/3) We can then substitute in our given values

g'(π/3) = (-3) (√(3)/2) + (4) (1/2)

g'(π/3) = -(3√3)/2 + 2 This gives our final value of g'(π/3)

To determine our value of h'(π/3), we will need to use the quotient rule instead.

Quotient Rule: Given two functions of x, u(x) and v(x), (u/v)' = (u'v - uv')/(v²)

Choose u(x) = cos(x), v(x) = f(x) --> Taking individual derivatives yields u'(x) = -sin(x), v'(x) = f'(x)

Note that based on our choices for u and v, h(x) = cos(x)/f(x) = u(x)/v(x). Therefore, we can directly apply our quotient rule:

h'(x) = (-sin(x)f(x) - cos(x)f'(x)) / (f(x)²) We can then plug x = π/3 into this formula

h'(π/3) = (-sin(π/3)f(π/3) - cos(π/3)f'(π/3)) / (f(π/3)²) Substitute in the proper values

h'(π/3) = (-((√3)/2) (4) - (1/2) (-3)) / (4)²)

h'(π/3) = (-2√3 + 3/2)/16

You can then simplify both answers as needed. I hope that this explanation is helpful!