If \(y = \log_e\left(x + \sqrt{x^{2} + 1}\right)\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{1}{\sqrt{x^{2} + 1}}\)
By the chain rule, the derivative of \(\log_e u\) is \(\frac{1}{u}\cdot\frac{du}{dx}\), with \(u = x + \sqrt{x^{2} + 1}\):
\( \displaystyle \frac{du}{dx} = 1 + \frac{2x}{2\sqrt{x^{2} + 1}} \)
\( \displaystyle = \frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1}} \)
So
\( \displaystyle \frac{dy}{dx} = \frac{1}{x + \sqrt{x^{2} + 1}} \cdot \frac{x + \sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}} \)
\( \displaystyle = \frac{1}{\sqrt{x^{2} + 1}} \)
The factor \(x + \sqrt{x^{2} + 1}\) cancels, which is what makes the answer so simple.