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To find the derivative of $\sqrt{3y-1}=5xy$ can I square both sides and then find the derivative? Would the following be correct?

$$\sqrt{3y-1}=5xy \leftrightarrow 3y-1=25x^2y^2$$ $$3y'=50xy^2+25x^2\cdot 2y\cdot y'$$ $$3y' - 50x^2\cdot y \cdot y' = 50xy^2$$ $$y'(3-50x^2y) = 50xy^2$$ $$y' = \frac{50xy^2}{3-50x^2y}$$

Edit: The textbook answer was $$ y' = \frac{10y\sqrt{3y-1}}{3-10x\sqrt{3y-1}} $$ I'm pretty sure I did something wrong?

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You should notice that the answer from your textbook is none other than substituting $ \sqrt{3y-1} $ for $ 5xy $ in your expression for $ y'=\frac{50xy^2}{3-50x^2y} $.

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