Question

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

Hint:

When simplifying the derivative of \( \log(\sqrt{x^2+1} - x) \), look for common terms in the numerator and denominator that can cancel each other out after a sign change.

Answer 1

Concept:
• Differentiate the function implicitly using the product rule.
• Use the chain rule for differentiating logarithmic and square root functions.
• Simplify the resulting expression to match the target equation.

Step 1: Differentiate both sides with respect to \( x \)
The equation is \( y\sqrt{x^2+1} = \log (\sqrt{x^2+1}-x) \).
Differentiating the left side using the product rule:
\[ \frac{d}{dx} (y\sqrt{x^2+1}) = \frac{dy}{dx}\sqrt{x^2+1} + y \cdot \frac{1}{2\sqrt{x^2+1}} \cdot 2x = \frac{dy}{dx}\sqrt{x^2+1} + \frac{xy}{\sqrt{x^2+1}} \]
Differentiating the right side using the chain rule:
\[ \frac{d}{dx} (\log (\sqrt{x^2+1}-x)) = \frac{1}{\sqrt{x^2+1}-x} \cdot \left( \frac{2x}{2\sqrt{x^2+1}} - 1 \right) = \frac{1}{\sqrt{x^2+1}-x} \cdot \frac{x - \sqrt{x^2+1}}{\sqrt{x^2+1}} \]

Step 2: Simplify the derivative on the right side
Factor out \( -1 \) in the numerator:
\[ \frac{d}{dx} (\log (\sqrt{x^2+1}-x)) = \frac{1}{\sqrt{x^2+1}-x} \cdot \frac{-(\sqrt{x^2+1} - x)}{\sqrt{x^2+1}} = -\frac{1}{\sqrt{x^2+1}} \]

Step 3: Equate the results and rearrange
Equating the expressions from Step 1 and
Step 2:
\[ \frac{dy}{dx}\sqrt{x^2+1} + \frac{xy}{\sqrt{x^2+1}} = -\frac{1}{\sqrt{x^2+1}} \]
Multiply the entire equation by \( \sqrt{x^2+1} \):
\[ \frac{dy}{dx}(x^2+1) + xy = -1 \]
Rearranging terms:
\[ (x^2+1) \frac{dy}{dx} + xy + 1 = 0 \]. Hence proved.