# How do you use the quotien rule to differentiate (3x^2+4) / sqrt(1+x^2)?

Oct 21, 2016

$f ' \left(x\right) = \frac{6 x \sqrt{1 + {x}^{2}} - \left(\frac{3 {x}^{3} + 4 x}{\sqrt{1 + {x}^{2}}}\right)}{1 + {x}^{2}}$

#### Explanation:

Quotient Rule

$f ' \left(x\right) = \frac{v u ' - u v '}{v} ^ 2$

$u = 3 {x}^{2} + 4$

$u ' = 6 x$

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

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

$f ' \left(x\right) = \frac{{\left(1 + {x}^{2}\right)}^{\frac{1}{2}} \cdot 6 x - \left(3 {x}^{2} + 4\right) \left(\frac{x}{\sqrt{1 + {x}^{2}}}\right)}{{\left(1 + {x}^{2}\right)}^{\frac{1}{2}}} ^ 2$

Simplify

$f ' \left(x\right) = \frac{6 x \sqrt{1 + {x}^{2}} - \left(\frac{x \left(3 {x}^{2} + 4\right)}{\sqrt{1 + {x}^{2}}}\right)}{1 + {x}^{2}}$

$f ' \left(x\right) = \frac{6 x \sqrt{1 + {x}^{2}} - \left(\frac{3 {x}^{3} + 4 x}{\sqrt{1 + {x}^{2}}}\right)}{1 + {x}^{2}}$

Common Denominator

$f ' \left(x\right) = \frac{6 x \sqrt{1 + {x}^{2}} \cdot \left(\frac{\sqrt{1 + {x}^{2}}}{\sqrt{1 + {x}^{2}}}\right) - \left(\frac{3 {x}^{3} + 4 x}{\sqrt{1 + {x}^{2}}}\right)}{1 + {x}^{2}}$

Simplify

$f ' \left(x\right) = \frac{\left(\frac{6 x \left(1 + {x}^{2}\right)}{\sqrt{1 + {x}^{2}}}\right) - \left(\frac{3 {x}^{3} + 4 x}{\sqrt{1 + {x}^{2}}}\right)}{1 + {x}^{2}}$

Distribute

$f ' \left(x\right) = \frac{\frac{6 x + 6 {x}^{3}}{\sqrt{1 + {x}^{2}}} - \left(\frac{3 {x}^{3} + 4 x}{\sqrt{1 + {x}^{2}}}\right)}{1 + {x}^{2}}$

Numerator simplified

$f ' \left(x\right) = \frac{\frac{6 x + 6 {x}^{3} - 3 {x}^{3} - 4 x}{\sqrt{1 + {x}^{2}}}}{1 + {x}^{2}}$

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

Multiply by the reciprocal

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

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

Simplify

$f ' \left(x\right) = \frac{2 x + 3 {x}^{3}}{{\left(1 + {x}^{2}\right)}^{\frac{1}{2}} \cdot {\left(1 + {x}^{2}\right)}^{\frac{2}{2}}}$

Simplify the denominator

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

Factor out an x from the numerator

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

Watch these examples of the quotient rule.