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:

Don't be intimidated by complex logs; the derivative of \(\log(\sqrt{x^2+1} \pm x)\) often simplifies to \( \pm 1/\sqrt{x^2+1} \).
Clearing radicals from the denominator as early as possible simplifies the subsequent algebra.

Answer 1

Concept:
• Implicit differentiation using the product rule: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \).
• Derivative of \(\log(f(x))\): \( \frac{f'(x)}{f(x)} \).
• Derivative of \(\sqrt{x^2+1}\): \( \frac{x}{\sqrt{x^2+1}} \).

Step 1: Differentiate both sides with respect to \( x \)
Using the product rule on the left-hand side: \[ y \cdot \frac{d}{dx}(\sqrt{x^2+1}) + \sqrt{x^2+1} \cdot \frac{dy}{dx} = \frac{d}{dx}[\log(\sqrt{x^2+1} - x)] \] \[ y \cdot \frac{2x}{2\sqrt{x^2+1}} + \sqrt{x^2+1} \frac{dy}{dx} = \frac{1}{\sqrt{x^2+1} - x} \cdot \left( \frac{x}{\sqrt{x^2+1}} - 1 \right) \]

Step 2: Simplify the right-hand side derivative
\[ \frac{xy}{\sqrt{x^2+1}} + \sqrt{x^2+1} \frac{dy}{dx} = \frac{1}{\sqrt{x^2+1} - x} \cdot \left( \frac{x - \sqrt{x^2+1}}{\sqrt{x^2+1}} \right) \] \[ \frac{xy}{\sqrt{x^2+1}} + \sqrt{x^2+1} \frac{dy}{dx} = \frac{-( \sqrt{x^2+1} - x)}{\sqrt{x^2+1}( \sqrt{x^2+1} - x)} \] \[ \frac{xy}{\sqrt{x^2+1}} + \sqrt{x^2+1} \frac{dy}{dx} = -\frac{1}{\sqrt{x^2+1}} \]

Step 3: Clear the denominator and rearrange
Multiply the entire equation by \( \sqrt{x^2+1} \): \[ xy + (\sqrt{x^2+1})^2 \frac{dy}{dx} = -1 \] \[ xy + (x^2+1) \frac{dy}{dx} = -1 \] \[ (x^2+1)\frac{dy}{dx} + xy + 1 = 0 \] Hence proved.