Question

Find the derivative of $y=\sqrt{x^2+1}$

• A. $\frac{x\sqrt{x^2+1}}{2}$
• B. $\frac{x}{\sqrt{x^2+1}}$
• C. $\frac{1}{2\sqrt{x^2+1}}$
• D. none of these

Answer 1

The correct option is B

$\frac{x}{\sqrt{x^2+1}}$

$y=\sqrt{x^2+1}$
$x^2+1=u....\left(1\right)$
$\Rightarrow \frac{du}{dx}=2x....\left(2\right)$
$\Rightarrow y=\sqrt{u}$
Now, $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$
=$\frac{1}{2\sqrt{u}}\times \frac{du}{dx}$
=$\frac{1}{2\sqrt{x^2+1}}\times 2x$ From (1) & (2)
$\frac{dy}{dx}=\frac{x}{\sqrt{x^2+1}}$