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Question Number 281 by samarth last updated on 25/Jan/15
If f(x)=tan^(−1) (√((1+sin x)/(1−sin x))), 0≤x≤π/2, then f ′(π/6)=?
$$\mathrm{If}\:{f}\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}},\:\mathrm{0}\leqslant{x}\leqslant\pi/\mathrm{2},\:\mathrm{then}\:{f}\:'\left(\pi/\mathrm{6}\right)=? \\ $$
Answered by 123456 last updated on 18/Dec/14
f(x)=tan^(−1) (√((1+sin x)/(1−sin x)))  (∂f/∂x)=(∂/∂x)(tan^(−1) (√((1+sin x)/(1−sin x))))  =(1/(1+((√((1+sin x)/(1−sin x))))^2 ))∙(∂/∂x)((√((1+sin x)/(1−sin x))))  =(1/(1+((1+sin x)/(1−sin x))))∙(1/(2(√((1+sin x)/(1−sin x)))))∙(∂/∂x)(((1+sin x)/(1−sin x)))  =(1/((1−sin x+1+sin x)/(1−sin x )))∙(1/2)(√(((1−sin x)/(1+sin x))∙))(((∂/∂x)(1+sin x)(1−sin x)−(1+sin x)(∂/∂x)(1−sin x))/((1−sin x)^2 ))  =((1−sin x)/2)∙(1/2)(√(((1−sin x)/(1+sin x))∙))((cos x(1−sin x)+cos x(1+sin x))/((1−sin x)^2 ))  =((1−sin x)/4)(√(((1−sin x)/(1+sin x))∙))((cos x−cos x sin x+cos x+cos x sin x)/((1−sin x)^2 ))  =((1−sin x)/4)∙(((1−sin x)^(1/2) )/((1+sin x)^(1/2) ))∙((2cos x)/((1−sin x)^2 ))  =(1/2)∙(((1−sin x)^(1/2) )/((1+sin x)^(1/2) ))∙((cos x)/(1−sin x))  x=(π/6)≡30°  =(1/2)∙(((1−sin (π/6))^(1/2) )/((1+sin (π/6))^(1/2) ))∙((cos (π/6))/(1−sin (π/6)))  =(1/2)∙(((1−(1/2))^(1/2) )/((1+(1/2))^(1/2) ))∙(((√3)/2)/(1−(1/2)))  =(1/2)∙((((1/2))^(1/2) )/(((3/2))^(1/2) ))∙(((√3)/2)/(1/2))  =(1/2)∙((1/( (√2)))/((√3)/( (√2))))∙((√3)/2)∙2  =(1/2)∙(1/( (√2)))∙((√2)/( (√3)))∙(√3)  =(1/2)
$${f}\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}} \\ $$$$\frac{\partial{f}}{\partial{x}}=\frac{\partial}{\partial{x}}\left(\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\left(\sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}}\right)^{\mathrm{2}} }\centerdot\frac{\partial}{\partial{x}}\left(\sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}}\centerdot\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}}}\centerdot\frac{\partial}{\partial{x}}\left(\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{1}−\mathrm{sin}\:{x}+\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}\:}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\centerdot}\frac{\frac{\partial}{\partial{x}}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)−\left(\mathrm{1}+\mathrm{sin}\:{x}\right)\frac{\partial}{\partial{x}}\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\centerdot}\frac{\mathrm{cos}\:{x}\left(\mathrm{1}−\mathrm{sin}\:{x}\right)+\mathrm{cos}\:{x}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{4}}\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}\centerdot}\frac{\mathrm{cos}\:{x}−\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}+\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{1}/\mathrm{2}} }\centerdot\frac{\mathrm{2cos}\:{x}}{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\left(\mathrm{1}−\mathrm{sin}\:{x}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\mathrm{1}/\mathrm{2}} }\centerdot\frac{\mathrm{cos}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}} \\ $$$${x}=\frac{\pi}{\mathrm{6}}\equiv\mathrm{30}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\left(\mathrm{1}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\mathrm{1}+\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right)^{\mathrm{1}/\mathrm{2}} }\centerdot\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{6}}}{\mathrm{1}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{2}} }\centerdot\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{2}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{2}} }\centerdot\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}}\centerdot\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\centerdot\mathrm{2} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\centerdot\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\centerdot\sqrt{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$