If \(y = \left(x + \sqrt{x^2 + 1}\right)^m\), then \(\displaystyle\frac{dy}{dx}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{my}{\sqrt{x^2 + 1}}\)
By the chain rule:
\(\displaystyle \frac{dy}{dx} = m\left(x + \sqrt{x^2 + 1}\right)^{m-1}\cdot\frac{d}{dx}\left(x + \sqrt{x^2 + 1}\right)\)
The derivative of the base is
\(\displaystyle \frac{d}{dx}\left(x + \sqrt{x^2 + 1}\right) = 1 + \frac{x}{\sqrt{x^2 + 1}}\)
\(\displaystyle = \frac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}\)
Its numerator is the base itself, so the power goes back up to m:
\(\displaystyle \frac{dy}{dx} = m\left(x + \sqrt{x^2 + 1}\right)^{m-1}\cdot\frac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}\)
\(\displaystyle = \frac{m\left(x + \sqrt{x^2 + 1}\right)^{m}}{\sqrt{x^2 + 1}}\)
\(\displaystyle = \frac{my}{\sqrt{x^2 + 1}}\)