If $y=\tan^{-1}\!\left(\dfrac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right),\ x^{2}\le 1$, then find $\dfrac{dy}{dx}$.

CBSE Class 12 · 2015 · DifferentiationLong Answer

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

, then find

\frac{d y}{d x}

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

Substitute

x^{2} = cos 2 θ

, so

1 + x^{2} = 2 cos θ, 1 - x^{2} = 2 sin θ

.
The argument becomes:

\frac{cos θ + sin θ}{cos θ - sin θ} = \frac{1 + tan θ}{1 - tan θ} = tan (\frac{π}{4} + θ)

Hence

y = \frac{π}{4} + θ = \frac{π}{4} + \frac{1}{2} cos^{- 1} (x^{2})

.
Differentiate:

\frac{d y}{d x} = \frac{1}{2} \cdot (- \frac{1}{1 - x ^{4}}) \cdot 2 x

\frac{d y}{d x} = - \frac{x}{1 - x ^{4}}

CBSE Class 12 Board · 2015 · Subjective