If $y\sqrt{x^{2}+1}=\log\!\left[\sqrt{x^{2}+1}-x\right]$, show that $(x^{2}+1)\dfrac{dy}{dx}+xy+1=0$.

CBSE Class 12 · 2006 · DifferentiationLong Answer

y x^{2} + 1 = lo g [x^{2} + 1 - x]

, show that

(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0

Answer: Proved

Differentiate both sides w.r.t.

x

. On the left use the product rule on

y x^{2} + 1

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

The right side simplifies, since

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

RHS = \frac{1}{x ^{2} + 1 - x} \cdot (- \frac{x ^{2} + 1 - x}{x ^{2} + 1}) = - \frac{1}{x ^{2} + 1}

So:

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

Multiply throughout by

x^{2} + 1

(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0

CBSE Class 12 Board · 2006 · Subjective