Question

Suppose f(π/6)=7 and f′(π/6)=−4 ,
and let g(x)=f(x)cosx and h(x)=sinx/f(x)

g'(π/6)=?
h'(π/6)=?

Answer 1

\[g(x)=f(x)\cos (x)\]
\[g'(x)=f'(x)\cos (x)-f(x)\sin (x)\]
\[g'(30)=f'(30)\cos(30)-f(30)\sin(30)\]

Answer 2

\[g'(30)=-4(\frac{\sqrt{3}}{2})-7(\frac{1}{2})\]

Answer 3

\[h(x)=\frac{\sin x}{f(x)}\]
\[h'(x)=\frac{f(x)\cos x-\sin xf'(x)}{[f(x)]^{2}}\
]\[h'(30)=\frac{f(30)\cos (30)-\sin (30)f'(30)}{[f(30]^{2}}\]

Answer 4

\[h'(30)=\frac{f(30)\cos (30)-\sin (30)f'(30)}{[f(30)]^{2}}\]

Answer 5

\[h'(30)=\frac{7(\frac{\sqrt{3}}{2}-\frac{1}{2}(-4)}{7^{2}}\]

Answer 6

\[h'(30)=\frac{14\sqrt{3}+2}{49}\]

Answer 7

I used 30 instead of pi/6 so I wouldn't have to type all those fractions.