# Derivative of $\sqrt{3y-1}=5xy$

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?

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