# How do you use the quotient rule to differentiate x/ sqrt (x^2 + 1)?

Oct 3, 2016

$\setminus \frac{1}{{\left({x}^{2} + 1\right)}^{\frac{3}{2}}}$

#### Explanation:

The quotient rule states that

$\frac{d}{\mathrm{dx}} \left(\frac{f}{g}\right) = \setminus \frac{f ' g - f g '}{{g}^{2}}$

Let's compute everything's required, and then we'll just substitute our results in this formula: we have

$f \left(x\right) = x \setminus \implies f ' \left(x\right) = 1$

$g \left(x\right) = \sqrt{{x}^{2} + 1} \setminus \implies g ' \left(x\right) = \frac{x}{\sqrt{{x}^{2} + 1}}$

Also, ${g}^{2} \left(x\right) = {x}^{2} + 1$

Thus,

$\setminus \frac{f ' g - f g '}{{g}^{2}} = \setminus \frac{1 \cdot \sqrt{{x}^{2} + 1} - x \cdot \frac{x}{\sqrt{{x}^{2} + 1}}}{{x}^{2} + 1}$

We can write this last result as

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

And then we can rearrange the numerator as

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

which equals

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