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

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Question Number 195137 by mathlove last updated on 25/Jul/23
f(x)=arctan(((4sinx)/(3+5cosx))) then f^′ ((π/3))=?
$${f}\left({x}\right)={arctan}\left(\frac{\mathrm{4}{sinx}}{\mathrm{3}+\mathrm{5}{cosx}}\right)\:\:\:{then}\:{f}^{'} \left(\frac{\pi}{\mathrm{3}}\right)=? \\ $$
Answered by Tokugami last updated on 02/Sep/23
f′(x)=(1/(1+(((4sin x)/(3+5cos x)))^2 )) (d/dx)(((4sin x)/(3+5cos x))) =(1/(1+((16sin^2 x)/((3+5cos x)^2 )))) (((3+5cos x)(4cos x)−(4sin x)(−5sin x))/((3+5cos x)^2 )) =((12cos x+20cos^2 x+20sin^2 x)/((3+5cos x)^2 +16sin^2 x)) =((12cos x+20)/(9+30cos x+25cos^2 x+16sin^2 x)) =((4(3cos x+5))/(25+30cos x+9cos^2 x))=((4(3cos x+5))/((3cos x+5)^2 )) =(4/(5+3cos x)) f′((π/3))=(4/(5+3cos((π/3))))=(4/(5+3((1/2))))=(4/(5+(3/2))) f′((π/3))=(8/(13))
$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{4sin}\:{x}}{\mathrm{3}+\mathrm{5cos}\:{x}}\right)^{\mathrm{2}} }\:\frac{{d}}{{dx}}\left(\frac{\mathrm{4sin}\:{x}}{\mathrm{3}+\mathrm{5cos}\:{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{16sin}^{\mathrm{2}} {x}}{\left(\mathrm{3}+\mathrm{5cos}\:{x}\right)^{\mathrm{2}} }}\:\frac{\left(\mathrm{3}+\mathrm{5cos}\:{x}\right)\left(\mathrm{4cos}\:{x}\right)−\left(\mathrm{4sin}\:{x}\right)\left(−\mathrm{5sin}\:{x}\right)}{\left(\mathrm{3}+\mathrm{5cos}\:{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{12cos}\:{x}+\mathrm{20cos}^{\mathrm{2}} {x}+\mathrm{20sin}^{\mathrm{2}} {x}}{\left(\mathrm{3}+\mathrm{5cos}\:{x}\right)^{\mathrm{2}} +\mathrm{16sin}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{12cos}\:{x}+\mathrm{20}}{\mathrm{9}+\mathrm{30cos}\:{x}+\mathrm{25cos}^{\mathrm{2}} {x}+\mathrm{16sin}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{4}\left(\mathrm{3cos}\:{x}+\mathrm{5}\right)}{\mathrm{25}+\mathrm{30cos}\:{x}+\mathrm{9cos}^{\mathrm{2}} {x}}=\frac{\mathrm{4}\cancel{\left(\mathrm{3cos}\:{x}+\mathrm{5}\right)}}{\left(\mathrm{3cos}\:{x}+\mathrm{5}\right)^{\cancel{\mathrm{2}}} } \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}+\mathrm{3cos}\:{x}} \\ $$$${f}'\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{4}}{\mathrm{5}+\mathrm{3cos}\left(\frac{\pi}{\mathrm{3}}\right)}=\frac{\mathrm{4}}{\mathrm{5}+\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{4}}{\mathrm{5}+\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${f}'\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{8}}{\mathrm{13}} \\ $$$$ \\ $$

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