y-1-4-ln-1-x-1-x-1-2-arctgx-y-

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

y = \frac{1}{4} l n \frac{1 + x}{1 - x} - \frac{1}{2} a r c t g x

y^{'} = ?

Commented by mohammad17 last updated on 29/Dec/20

y^{^{'}} = \frac{1}{4} (\frac{[(1 - x) (1) - (1 + x) (- 1)] (1 + x)}{{(1 - x)}^{3}}) - \frac{1}{2 x^{2} + 2}

y^{^{'}} = \frac{2 x + 2}{{(1 - x)}^{3}} - \frac{1}{2 x^{2} + 2}

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

y = \frac{1}{4} (l o g (1 + x)) - \frac{1}{4} l o g (1 - x) - \frac{1}{2} {t a n}^{- 1} x

y^{'} = \frac{1}{4 (1 + x)} + \frac{1}{4 (1 - x)} - \frac{1}{2} . \frac{1}{1 + x^{2}} = \frac{x^{2}}{1 - x^{4}}

Answered by hknkrc46 last updated on 29/Dec/20

∙ \frac{d}{d x} \ln \frac{a (x)}{b (x)} = \frac{a^{^{'}} (x)}{a (x)} - \frac{b^{^{'}} (x)}{b (x)}

✓ \frac{d}{d x} \ln \frac{1 + x}{1 - x} = \frac{{(1 + x)}^{^{'}}}{1 + x} - \frac{{(1 - x)}^{'}}{1 - x}

= \frac{1}{1 + x} + \frac{1}{1 - x}

∙ \frac{d}{d x} arctg x = \frac{1}{1 + x^{2}}

★ y^{^{'}} = \frac{1}{4} (\frac{1}{1 + x} + \frac{1}{1 - x}) - \frac{1}{2 (1 + x^{2})}

= \frac{1}{2} (\frac{1}{1 - x^{2}} - \frac{1}{1 + x^{2}}) = \frac{x^{2}}{1 - x^{4}}

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