Respuesta :
[tex]\bf g(x)=f(x)sin(x)\implies \cfrac{dg}{dx}=\cfrac{df}{dx}sin(x)+f(x)cos(x)
\\\\\\
g'\left( \frac{\pi }{3} \right)=f'\left( \frac{\pi }{3} \right)sin\left( \frac{\pi }{3} \right)+f\left( \frac{\pi }{3} \right)cos\left( \frac{\pi }{3} \right)
\\\\\\
g'\left( \frac{\pi }{3} \right)=-5\cdot \cfrac{\sqrt{3}}{2}+3\cdot \cfrac{1}{2}\implies \boxed{g'\left( \frac{\pi }{3} \right)=\cfrac{3-5\sqrt{3}}{2}}\\\\
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[/tex]
[tex]\bf h(x)=\cfrac{cos(x)}{f(x)}\implies \cfrac{dh}{dx}=\cfrac{-sin(x)f(x)-cos(x)\frac{df}{dx}}{\left[ f(x) \right]^2} \\\\\\ h'\left( \frac{\pi }{3} \right)=\cfrac{-sin\left( \frac{\pi }{3} \right)f\left( \frac{\pi }{3} \right)-cos\left( \frac{\pi }{3} \right)f'\left( \frac{\pi }{3} \right)}{\left[ f\left( \frac{\pi }{3} \right) \right]^2}[/tex]
[tex]\bf h'\left( \frac{\pi }{3} \right)=\cfrac{-\frac{\sqrt{3}}{2}\cdot 3-\frac{1}{2}\cdot -5}{3^2}\implies h'\left( \frac{\pi }{3} \right)=\cfrac{\frac{5-3\sqrt{3}}{2}}{9}\implies \boxed{h'\left( \frac{\pi }{3} \right)=\cfrac{5-3\sqrt{3}}{18}}[/tex]
[tex]\bf h(x)=\cfrac{cos(x)}{f(x)}\implies \cfrac{dh}{dx}=\cfrac{-sin(x)f(x)-cos(x)\frac{df}{dx}}{\left[ f(x) \right]^2} \\\\\\ h'\left( \frac{\pi }{3} \right)=\cfrac{-sin\left( \frac{\pi }{3} \right)f\left( \frac{\pi }{3} \right)-cos\left( \frac{\pi }{3} \right)f'\left( \frac{\pi }{3} \right)}{\left[ f\left( \frac{\pi }{3} \right) \right]^2}[/tex]
[tex]\bf h'\left( \frac{\pi }{3} \right)=\cfrac{-\frac{\sqrt{3}}{2}\cdot 3-\frac{1}{2}\cdot -5}{3^2}\implies h'\left( \frac{\pi }{3} \right)=\cfrac{\frac{5-3\sqrt{3}}{2}}{9}\implies \boxed{h'\left( \frac{\pi }{3} \right)=\cfrac{5-3\sqrt{3}}{18}}[/tex]
