Find $\dfrac{dy}{dx}$, if $y=\sin^{-1}\!\left[x\sqrt{1-x}-\sqrt{x}\,\sqrt{1-x^{2}}\right]$.

$\dfrac{1}{\sqrt{1-x^{2}}}-\dfrac{1}{2\sqrt{x(1-x)}}$

CBSE Class 12 · 2010 · DifferentiationLong Answer

Find

\frac{d y}{d x}

, if

y = sin^{- 1} [x 1 - x - x 1 - x^{2}]

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

Recognise the form

sin^{- 1} [a 1 - b^{2} - b 1 - a^{2}] = sin^{- 1} a - sin^{- 1} b

with

a = x, b = x

(note

1 - (x)^{2} = 1 - x

y = sin^{- 1} x - sin^{- 1} x

Differentiate:

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

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

CBSE Class 12 Board · 2010 · Subjective