If y = `log ((1 - x^2)/(1 + x^2))`, then `"dy"/"dx"` is equal to ______.

If y = `log ((1 - x^2)/(1 + x^2))`, then `"dy"/"dx"` is equal to `(-4x)/(1 - x^4)`. Explanation: Given that: y = `log ((1 - x^2)/(1 + x^2))` ⇒ y = log(1 – x2) – log(1 + x2) ....`[because log x/y = log x - log y]` Differentiating both sides w.r.t. x `"dy"/"dx" = 1/(1 - x^2) * "d"/"dx"(1 - x^2) - 1/(1 + x^2) (1 + x^2)` = `(-2x)/(1 - x^2) - (2x)/(1 + x^2)` = `(-2x - 2x^3 - 2x + 2x^3)/((1 - x^2)(1 + x^2))` = `(-4x)/(1 - x^4)`.

If y = log(1-x21+x2), then dydxdydx is equal to ______. - Mathematics

Question

If y = $\log (\frac{1 - x^{2}}{1 + x^{2}})$ , then $\frac{dy}{dx}$ is equal to ______.

Options

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

MCQ

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Solution

If y = $\log (\frac{1 - x^{2}}{1 + x^{2}})$ , then $\frac{dy}{dx}$ is equal to $\frac{- 4 x}{1 - x^{4}}$ .

Explanation:

Given that: y = $\log (\frac{1 - x^{2}}{1 + x^{2}})$

⇒ y = log(1 – x²) – log(1 + x²) .... $[∵ \log \frac{x}{y} = \log x - \log y]$

Differentiating both sides w.r.t. x

$\frac{dy}{dx} = \frac{1}{1 - x^{2}} \cdot \frac{d}{dx} (1 - x^{2}) - \frac{1}{1 + x^{2}} (1 + x^{2})$

= $\frac{- 2 x}{1 - x^{2}} - \frac{2 x}{1 + x^{2}}$

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

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

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Logarithmic Differentiation

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Q 91 Q 90 Q 92

Chapter 5: Continuity And Differentiability - Exercise [Page 114]

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Chapter 5 Continuity And Differentiability
Exercise | Q 91 | Page 114

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