If y = log left(frac{1-x^{2}}{1+x^{2}}right), | vedclass

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(A) Given $y = \log \left(\frac{1-x^{2}}{1+x^{2}}\right)$.Using the chain rule and the quotient rule for differentiation:$\frac{dy}{dx} = \frac{1}{\fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $y = \log \left(\frac{1-x^{2}}{1+x^{2}}\right)$.Using the chain rule and the quotient rule for differentiation:$\frac{dy}{dx} = \frac{1}{\frac{1-x^{2}}{1+x^{2}}} \cdot \frac{d}{dx} \left(\frac{1-x^{2}}{1+x^{2}}\right)$$\frac{dy}{dx} = \frac{1+x^{2}}{1-x^{2}} \cdot \left[ \frac{(-2x)(1+x^{2}) - (2x)(1-x^{2})}{(1+x^{2})^{2}} \right]$$\frac{dy}{dx} = \frac{1}{1-x^{2}} \cdot \left[ \frac{-2x - 2x^{3} - 2x + 2x^{3}}{1+x^{2}} \right]$$\frac{dy}{dx} = \frac{1}{1-x^{2}} \cdot \left[ \frac{-4x}{1+x^{2}} \right]$$\frac{dy}{dx} = \frac{-4x}{(1-x^{2})(1+x^{2})}$$\frac{dy}{dx} = \frac{-4x}{1-x^{4}}$

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

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$\frac{d}{dx} [\log(\cos x)]$

$\frac{d}{dx} \left[ \log \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right] = $

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