PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for If Y = Tan − 1 ( 2 X 1 − X 2 ) + Sec − 1 ( 1 + X 2 1 − X 2 ) , X > 0 ,Prove that D Y D X = 4 1 + X 2 ? - PUC Karnataka Science Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

If $y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x > 0$ ,prove that $\frac{dy}{dx} = \frac{4}{1 + x^2}$ ?

#### Solution

$\text{ Here, y} = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + se c^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right)$
$\Rightarrow y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right)$
$\text{Put x} = \tan\theta$
$\therefore y = \tan^{- 1} \left( \frac{2\tan\theta}{1 - \tan^2 \theta} \right) + \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right)$
$\Rightarrow y = \tan^{- 1} \left( \tan 2\theta \right) + \cos^{- 1} \left( \cos 2\theta \right)$
$\Rightarrow y = 2\theta + 2\theta$
$\Rightarrow y = 4\theta$
$\Rightarrow y = 4 \tan^{- 1} x \left[ \text{using, x} = tan\theta \right]$

Differentiate it with respect to x,

$\therefore \frac{d y}{d x} = \frac{4}{1 + x^2}$

Is there an error in this question or solution?

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Solution If Y = Tan − 1 ( 2 X 1 − X 2 ) + Sec − 1 ( 1 + X 2 1 − X 2 ) , X > 0 ,Prove that D Y D X = 4 1 + X 2 ? Concept: Simple Problems on Applications of Derivatives.
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