f-x-arctan-4sinx-3-5cosx-then-f-pi-3-

Question Number 195137 by mathlove last updated on 25/Jul/23

f (x) = a r c t a n (\frac{4 s i n x}{3 + 5 c o s x}) t h e n f^{^{'}} (\frac{π}{3}) = ?

Answered by Tokugami last updated on 02/Sep/23

f^{'} (x) = \frac{1}{1 + {(\frac{4 \sin x}{3 + 5 \cos x})}^{2}} \frac{d}{d x} (\frac{4 \sin x}{3 + 5 \cos x})

= \frac{1}{1 + \frac{{16 \sin}^{2} x}{{(3 + 5 \cos x)}^{2}}} \frac{(3 + 5 \cos x) (4 \cos x) - (4 \sin x) (- 5 \sin x)}{{(3 + 5 \cos x)}^{2}}

= \frac{12 \cos x + {20 \cos}^{2} x + {20 \sin}^{2} x}{{(3 + 5 \cos x)}^{2} + {16 \sin}^{2} x}

= \frac{12 \cos x + 20}{9 + 30 \cos x + {25 \cos}^{2} x + {16 \sin}^{2} x}

= \frac{4 (3 \cos x + 5)}{25 + 30 \cos x + {9 \cos}^{2} x} = \frac{4 (3 \cos x + 5)}{{(3 \cos x + 5)}^{2}}

= \frac{4}{5 + 3 \cos x}

f^{'} (\frac{π}{3}) = \frac{4}{5 + 3 \cos (\frac{π}{3})} = \frac{4}{5 + 3 (\frac{1}{2})} = \frac{4}{5 + \frac{3}{2}}

f^{'} (\frac{π}{3}) = \frac{8}{13}

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