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Question Number 127434 by MathSh last updated on 29/Dec/20

y=(1/4)ln((1+x)/(1−x))−(1/2)arctgx y′=?

$${y}=\frac{\mathrm{1}}{\mathrm{4}}{ln}\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{2}}{arctgx} \\ $$$${y}'=? \\ $$

Commented by mohammad17 last updated on 29/Dec/20

y^′ =(1/4)((([(1−x)(1)−(1+x)(−1)](1+x))/((1−x)^3 )))−(1/(2x^2 +2)) y^′ =((2x+2)/((1−x)^3 ))−(1/(2x^2 +2))

$${y}^{'} =\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\left[\left(\mathrm{1}−{x}\right)\left(\mathrm{1}\right)−\left(\mathrm{1}+{x}\right)\left(−\mathrm{1}\right)\right]\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\right)−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}} \\ $$$$ \\ $$$${y}^{'} =\frac{\mathrm{2}{x}+\mathrm{2}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}} \\ $$

Answered by Dwaipayan Shikari last updated on 29/Dec/20

y=(1/4)(log(1+x))−(1/4)log(1−x)−(1/2)tan^(−1) x y′=(1/(4(1+x)))+(1/(4(1−x)))−(1/2).(1/(1+x^2 ))=(x^2 /(1−x^4 ))

$${y}=\frac{\mathrm{1}}{\mathrm{4}}\left({log}\left(\mathrm{1}+{x}\right)\right)−\frac{\mathrm{1}}{\mathrm{4}}{log}\left(\mathrm{1}−{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} {x} \\ $$$${y}'=\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{x}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} } \\ $$

Answered by hknkrc46 last updated on 29/Dec/20

• (d /dx)ln ((a(x))/(b(x))) = ((a^′ (x))/(a(x))) − ((b^′ (x))/(b(x))) ✓ (d/dx)ln ((1 + x)/(1 − x)) = (((1 + x)^′ )/(1 + x)) − (((1 − x)′)/(1−x)) = (1/(1 + x)) + (1/(1−x)) • (d/dx)arctg x = (1/(1 + x^2 )) ★ y^′ = (1/4)((1/(1 + x)) + (1/(1−x))) − (1/(2(1 + x^2 ))) = (1/2)((1/(1 − x^2 )) − (1/(1 + x^2 ))) = (x^2 /(1 − x^4 ))

$$\bullet\:\frac{{d}\:}{{dx}}\mathrm{ln}\:\frac{{a}\left({x}\right)}{{b}\left({x}\right)}\:=\:\frac{{a}^{'} \left({x}\right)}{{a}\left({x}\right)}\:−\:\frac{{b}^{'} \left({x}\right)}{{b}\left({x}\right)} \\ $$$$\checkmark\:\frac{{d}}{{dx}}\mathrm{ln}\:\frac{\mathrm{1}\:+\:{x}}{\mathrm{1}\:−\:{x}}\:=\:\frac{\left(\mathrm{1}\:+\:{x}\right)^{'} }{\mathrm{1}\:+\:{x}}\:−\:\frac{\left(\mathrm{1}\:−\:{x}\right)'}{\mathrm{1}−{x}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\bullet\:\frac{{d}}{{dx}}\mathrm{arctg}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} } \\ $$$$\bigstar\:{y}^{'} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{1}\:+\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}\:−\:{x}^{\mathrm{4}} } \\ $$

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