# How do you find the derivative of sqrt(x^2+1)?

Nov 1, 2016

$y = \sqrt{{x}^{2} + 1}$

${y}^{2} = {x}^{2} + 1$

Use implicit differentiation on the left hand side of the equation and ordinary differentiation on the right hand side of the equation.

$2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

$y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{\sqrt{{x}^{2} + 1}}$

Nov 1, 2016

Use the chain rule and the power rule.

#### Explanation:

The power rule says that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$

The chain rule, when combined with the power rule (sometimes called "the general power rule" says that

$\frac{d}{\mathrm{dx}} \left({u}^{n}\right) = n {u}^{n - 1} \frac{\mathrm{du}}{\mathrm{dx}}$

So

$\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 1}\right) = \frac{d}{\mathrm{dx}} \left({\left({x}^{2} + 1\right)}^{\frac{1}{2}}\right)$

$= \frac{1}{2} {\left({x}^{2} + 1\right)}^{\frac{1}{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)$

$= \frac{1}{2} {\left({x}^{2} + 1\right)}^{- \frac{1}{2}} \cdot \left(2 x\right)$

$= \frac{x}{{x}^{2} + 1} ^ \left(\frac{1}{2}\right)$

 = x/(sqrt(x^2+1)

The differentiation is sped up considerably by learning the $\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) = \frac{1}{2 \sqrt{x}}$.

After this is learned, we can simply write

$\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 1}\right) = \frac{1}{2 \sqrt{{x}^{2} + 1}} \cdot 2 x = \frac{x}{\sqrt{{x}^{2} + 1}}$