# How do you differentiate x^2(3x+1)^(1/2)?

May 19, 2015

The answer is : $f ' \left(x\right) = \frac{x \left(15 x + 4\right)}{2 \sqrt{3 x + 1}}$.

Let's have a quick look at your function :

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

The derivative of such a form is given by the product rule :

$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

The derivative of $g \left(x\right) = {x}^{2}$ is $g ' \left(x\right) = 2 x$

and the derivative of $h \left(x\right) = {\left(3 x + 1\right)}^{\frac{1}{2}}$ is

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

Therefore, the derivative of $f \left(x\right)$ is :

$f ' \left(x\right) = 2 x \cdot {\left(3 x + 1\right)}^{\frac{1}{2}} + {x}^{2} \cdot \frac{3}{2 {\left(3 x + 1\right)}^{\frac{1}{2}}} = \frac{4 x \cdot {\left({\left(3 x + 1\right)}^{\frac{1}{2}}\right)}^{2} + 3 {x}^{2}}{2 {\left(3 x + 1\right)}^{\frac{1}{2}}} = \frac{4 x \left(3 x + 1\right) + 3 {x}^{2}}{2 {\left(3 x + 1\right)}^{\frac{1}{2}}} = \frac{15 {x}^{2} + 4 x}{2 {\left(3 x + 1\right)}^{\frac{1}{2}}} = \frac{x \left(15 x + 4\right)}{2 \sqrt{3 x + 1}}$.

That's it.