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

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  • $\begingroup$ The set up is wrong its y = 2x^2 + 6 square root x divided by 9x $\endgroup$
    – user137452
    Oct 14, 2014 at 19:01

5 Answers 5

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

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

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  • $\begingroup$ That doesn't really state anything; it's just an expression. $\endgroup$
    – beep-boop
    Oct 15, 2014 at 12:13
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$$y = 2x^2 + \frac{6\sqrt x}{9x} = 2x^2 +\frac {2}{3\sqrt x} = 2x^2 + \frac 23\cdot x^{-1/2}$$

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

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 $$\begin{align} \frac{dx}{dy} & = \frac 29 - \frac 12\cdot\frac 23 \cdot x^{-3/2}\\ \\ & = \frac 29 - \frac{1}{3x^{3/2}}\\ \\ & = \frac{2x^{3/2} - 3}{9x^{3/2}} \end{align}$$

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  • $\begingroup$ The set up is wrong its y = 2x^2 + 6 square root x divided by 9x $\endgroup$
    – user137452
    Oct 14, 2014 at 19:00
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    $\begingroup$ Please see my edit. $\endgroup$
    – amWhy
    Oct 14, 2014 at 19:15
  • $\begingroup$ So was my answer right? $\endgroup$
    – user137452
    Oct 14, 2014 at 20:12
  • $\begingroup$ No, the answer is what you see in my post. $$y' = \frac 29 - \frac{1}{3x^{3/2}} = \frac{2x^{3/2} - 3}{9x^{3/2}}$$ $\endgroup$
    – amWhy
    Oct 14, 2014 at 20:50
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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 $$4x-\frac{1}{3}\times x^{-\frac{3}{2}}$$

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

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

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