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Suppose that f(pi/3)=4 and f'(pi/3)=-2, and let g(x)= (cosx)/f(x). find g'(x) (pi/3)

User Zdf
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g(x)=(\cos x)/(f(x))

\implies g'(x)=-(f(x)\sin x+f'(x)\cos x)/(f(x)^2)

You know that
f\left(\frac\pi3\right)=4 and
f'\left(\frac\pi3\right)=-2, so


g'\left(\frac\pi3\right)=-(4\sin\frac\pi3-2\cos\frac\pi3)/(4^2)=(1-2\sqrt3)/(16)
User Samidamaru
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