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If $y = \cos^{-1}\left (\frac{1-x^{2}}{1+x^{2}} \right )$ then $\frac{\mathrm{d} y}{\mathrm{d} x} =$ ? $(x > 0)$

Option 1)
$\frac{1}{1 + x^{2}}$

Option 2)
$\frac{2}{1 + x^{2}}$

Option 3)
$\frac{3}{1 + x^{2}}$

Option 4)
$-\frac{2}{1 + x^{2}}$

Answers (1)

best_answer

As we have learnt,

Differentiation of inverse trigonometric function -

Let

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

$put\:\:x=tan\theta$

$\therefore \:\:sin^{-1}(\frac{2tan\theta}{1+tan^{2}\theta})$

$=sin^{-1}sin(2\theta)$

$=2\theta=2tan^{-1}x$

$\therefore \:\:\frac{dy}{dx}=\frac{2}{1+x^{2}}$

- wherein

Use : for

$\frac{2x}{1+x^{2}},\:\:\frac{1-x^{2}}{1+x^{2}}\:\:and\:\:\frac{2x}{1-x^{2}}$

$put\:\:x=tan\theta\:\:and\:gives:\:sin2\theta$

$cos2\theta\:\:and\;\;tan2\theta\:\:respectively$

Let $x = \tan\theta$

$\\*\Rightarrow y = \cos^{-1}\left (\frac{1 - \tan^2\theta}{1 + \tan^2\theta} \right ) = \cos^{-1}(\cos2\theta) = 2\theta = 2\tan^{-1}x \\* \therefore\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(2\tan^{-1} x) = \frac{2}{1+x^2}$

Option 1)

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

Option 2)

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

Option 3)

$\frac{3}{1 + x^{2}}$

Option 4)

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

Posted by

Himanshu

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