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

Jan 26, 2017

$\frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 1}}\right) = \frac{1}{\left({x}^{2} + 1\right) \sqrt{{x}^{2} + 1}} = \frac{1}{{x}^{2} + 1} ^ \left(\frac{3}{2}\right)$

Explanation:

Using the quotient rule:

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

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

We can now use the chain rule to calculate:

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

and substitute it in the expression above:

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

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

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