Suppose f(π/3) = 2 and f '(π/3) = −5, and let g(x) = f(x) sin(x) and h(x) = cos(x)/f(x).

Question

Suppose f(π/3) = 2 and f '(π/3) = −5,

 and let g(x) = f(x) sin(x)

 and h(x) = cos(x)/f(x).

 Find the following.

(a)    g'(π/3)

(b)    h'(π/3)

Answer 1

(a) g'(x) = f'(x)sinx + f(x)cosx; g'(π/3) = f'(π/3)sin(π/3) + f(π/3)cos(π/3) = - 5·√3/2 + 2·1/2 = -5√3/2 + 1

(b) h'(x) = (- sinxf(x) - cosxf'(x))/f(x)²; h'(π/3) = (- √3/2·2 + 1/2·5)/4 = (5 - 2√3)/8

Answer 2

The problem tests your ability to compute derivatives using the product and quotient rules. It is important to note that in all of the functions, f(x), g(x), and h(x), there is only one variable, namely x. So we don't have to worry about applying the chain rule in this problem and can carry on as we normally would.

A) The first step would be to compute g'(x). We apply the product rule. (Remember: (1)d2 + (2)d1. We take the first term multiplied by the derivative of the second add the product of the second term with the derivative of the first term.)

g'(x) = f(x)cos(x) + sin(x)f'(x)

To calculate g'(π/3), we just have to plug in π/3 for x. The values for f(π/3) and f'(π/3) are given in the problem.

B) We approach this part in the same way as A, except we will use the quotient rule. (Remember: Low dHigh minus High dLow, square the bottom, and off we go!)

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

Now, we just need to substitute π/3 for x and solve.