\( \displaystyle\frac{d}{dx}\log\left(x+\sqrt{x^2+4}\right) \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{\sqrt{x^2+4}} \)
By the chain rule, the derivative of log u is \( \frac{1}{u}\cdot\frac{du}{dx} \).
\( \displaystyle \frac{d}{dx}\left(x+\sqrt{x^2+4}\right)=1+\frac{2x}{2\sqrt{x^2+4}} \)
\( \displaystyle =\frac{\sqrt{x^2+4}+x}{\sqrt{x^2+4}} \)
\( \displaystyle \frac{d}{dx}\log\left(x+\sqrt{x^2+4}\right) \)
\( \displaystyle =\frac{1}{x+\sqrt{x^2+4}}\cdot\frac{x+\sqrt{x^2+4}}{\sqrt{x^2+4}} \)
\( \displaystyle =\frac{1}{\sqrt{x^2+4}} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{x+\sqrt{x^2+4}} \) — Incorrect. This stops at \( \frac{1}{u} \) and leaves out the factor \( \frac{du}{dx} \) of the chain rule.
Option (2) \( \displaystyle\frac{2x}{x+\sqrt{x^2+4}} \) — Incorrect. The derivative of \( x+\sqrt{x^2+4} \) is \( \frac{x+\sqrt{x^2+4}}{\sqrt{x^2+4}} \), not 2x.
Option (3) \( \displaystyle\frac{2x}{x-\sqrt{x^2+4}} \) — Incorrect. At x = 0 this option gives 0, while the derivative is \( \frac{1}{\sqrt{4}}=\frac{1}{2} \).
Tip: Remember \( \frac{d}{dx}\log\left(x+\sqrt{x^2+a^2}\right)=\frac{1}{\sqrt{x^2+a^2}} \); it is the reverse of the standard integral \( \int\frac{dx}{\sqrt{x^2+a^2}} \).