# Differentiate the following function: $y = \frac{2x^2 + 6\sqrt{x} }{9x}$

My answer is $- \dfrac{1}{9 \sqrt{x} }$, however, the program I am using states that I am wrong. Where have I went wrong?

• The set up is wrong its y = 2x^2 + 6 square root x divided by 9x Oct 14, 2014 at 19:01

$y=2x^2+\frac{2}{3}x^{-1/2}$. Now use the power rule to differentiate.

$4x-\frac{2}{3}\cdot \frac{1}{2}x^{-\frac{3}{2}}$

• That doesn't really state anything; it's just an expression. Oct 15, 2014 at 12:13

$$y = 2x^2 + \frac{6\sqrt x}{9x} = 2x^2 +\frac {2}{3\sqrt x} = 2x^2 + \frac 23\cdot x^{-1/2}$$

$$y' = 4x - \frac 12\cdot\frac 23 \cdot x^{-3/2} = 4x - \frac{1}{3x^{3/2}}$$

EDIT:

Given the OP's comments and attempted edit, it seems that the function of interest is intended to be $$y = \frac{2x^2 + 6\sqrt{x}}{9x}= \frac 29 x + \frac 23\cdot x^{-1/2}$$ If so, then \begin{align} \frac{dx}{dy} & = \frac 29 - \frac 12\cdot\frac 23 \cdot x^{-3/2}\\ \\ & = \frac 29 - \frac{1}{3x^{3/2}}\\ \\ & = \frac{2x^{3/2} - 3}{9x^{3/2}} \end{align}

• The set up is wrong its y = 2x^2 + 6 square root x divided by 9x Oct 14, 2014 at 19:00
• Please see my edit. Oct 14, 2014 at 19:15
• So was my answer right? Oct 14, 2014 at 20:12
• No, the answer is what you see in my post. $$y' = \frac 29 - \frac{1}{3x^{3/2}} = \frac{2x^{3/2} - 3}{9x^{3/2}}$$ Oct 14, 2014 at 20:50

The derivative of your first term is just $4x$ and if you use the power rule for the second one, you get $-\frac{1}{3}\times x^{-\frac{3}{2}}$ so eventually you get $$4x-\frac{1}{3}\times x^{-\frac{3}{2}}$$

$y = \frac{2x^{2}}{9x} + \frac{6x^\frac{1}{2}}{9x}=\frac{2x}{9} + \frac{2}{3}x^{-\frac{1}{2}}$.

Then $\frac{dy}{dx}=\frac{2}{9} -\frac{2}{3}⋅-\frac{1}{2}x^{-\frac{3}{2}}$