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given the equation $\sqrt{x+3y}-4xy+3y=5$, solve for $y'$

my answer : $y' = \frac{-1+32xy^2-18y}{3-32x^2}$

I feel like my answer is wrong.

The first thing I did was square both sides because trying to isolate y' in one side was difficult when I have a square root. $\frac{1}{2}(x+3y)^\frac{-1}{2}(1 + 3y')-4xy'-4y+3y'= 0 $ (unless I did something terribly wrong)

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Differentiate both sides and you get:

$$\frac{1+3y'}{2\sqrt{x+3y}}-4y-4xy'+3y'=0$$ Collecting the terms with $y'$ you get:

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

You can substitute $\sqrt{x+3y}$ with $5+4xy-3y$ to get:

$$y'\left ( \frac{3}{10+8xy-6y}-4x+3\right )=4y-\frac{1}{10+8xy-6y}$$

Simplifying you get:

$$y'=\frac{32xy^2-24y^2+40y-1}{-32x^2y+48xy-40x-18y+33}$$

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it is $$\frac{1}{2\sqrt{x+3y}}-4y+y'\left(\frac{3}{2\sqrt{x+3y}}-4x+3\right)=0$$ for $$\sqrt{x+3y}$$ you can substitute $$\sqrt{x+3y}=5-3y+4xy$$

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  • $\begingroup$ Shouldn't it be $-4x$ and not $-4y$ (in the parentheses)? $\endgroup$ – cansomeonehelpmeout Feb 18 '18 at 13:39
  • $\begingroup$ yes it must be $$-4x$$ thank you very much! $\endgroup$ – Dr. Sonnhard Graubner Feb 18 '18 at 13:42

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