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Algebra of Derivatives

If y =log1- x...

Question

If $y = l o g (\frac{1 - x^{2}}{1 + x^{2}}), t h e n \frac{d y}{d x}$ then dy/dx is equal to……

A

$\frac{- 4 x}{1 - x^{4}}$

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B

$\frac{- 4 x^{3}}{1 - x^{4}}$

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C

$\frac{1}{4 - x^{4}}$

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D

$\frac{- 4 x^{3}}{1 - x^{4}}$

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Solution

The correct option is A
$\frac{- 4 x}{1 - x^{4}}$

$\frac{d y}{d x} = \frac{1 + x^{2}}{1 - x^{2}} \times \frac{- 2 x (1 + x^{2}) - (1 - x^{2})}{(1 + x^{2})^{2}} = - \frac{- 4 x}{1 - x^{4}}$

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Similar questions

y = log (\frac{1 - x^{2}}{1 + x^{2}}),

\frac{d y}{d x}

y = {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) + {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) - {tan}^{1} (\frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x^{4}})

\frac{d y}{d x} = \frac{1}{1 + x^{2}}

If $y = l o g (\frac{1 - x^{2}}{1 + x^{2}}), t h e n \frac{d y}{d x}$ then dy/dx is equal to……

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

\frac{d y}{d x}

is equal to ___

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| x | < 1

\frac{d y}{d x}

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