Differentiate the following with respect to $x$: $\tan^{-1}\!\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$.

CBSE Class 12 · 2008 · DifferentiationLong Answer

Differentiate the following with respect to

x

tan^{- 1} (\frac{1 + x - 1 - x}{1 + x + 1 - x})

Answer: $\frac{1}{2 1 - x ^{2}}$

Substitute

x = cos 2 θ

, so

1 + x = 2 cos θ, 1 - x = 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 = tan^{- 1} [tan (\frac{π}{4} - θ)] = \frac{π}{4} - θ = \frac{π}{4} - \frac{1}{2} cos^{- 1} x

.
Differentiate:

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

\frac{d y}{d x} = \frac{1}{2 1 - x ^{2}}

CBSE Class 12 Board · 2008 · Subjective