Question

Suppose f(π/3) = 3 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

for g' I thought the answer would be 7/2 - 5 sqrt(3/2) but this seems to be incorrect. Any input why?

Answer 2

since g(x) = f(x)sinx, then g'(x) = f'(x)sinx + f(x)[sinx]', by the product rule.

so
g'(pi/3) = f'(pi/3) sin(pi/3) + f(pi/3) cos(pi/3)

Answer 3

for h', the answer is 1/9 (5/2 - (3 sqrt 3)/2 No clue on g'

Answer 4

g'(pi/3) = f'(pi/3) sin(pi/3) + f(pi/3) cos(pi/3)

= \(\large (-5)(\frac{\sqrt 3}{2})+(3)(\frac{1}{2}) \)

can u take it from here???

Answer 5

byteme: Yeah, for g' I get to 3 (1/2) + sqrt(3/2) (-5), since those were the given values for those. the result I get is 7/2 - 5 sqrt (3/2) but this is wrong.

Answer 6

hm I should try and make better looking equations like those heh

Answer 7

byteme: I get \[7\div2 - 5 \sqrt{3 \div 2}\] as the result. This is incorrect, so if you have a better answer or an explanation of why its incorrect, I'd appreciate it. That being said, I'd appreciate the answer from anyone.

Answer 8

anyone?