Implicit Differentiation: x^3 + sqrt[2xy] + y^2 = 12

[Sophie]

Please could someone check if I have done the following correctly, Thanks:

Use implicit differentiation to find dy/dx.


\(\displaystyle \L\\\begin{array}{l}
x^3 + \sqrt {2xy} + y^2 = 12 \\

\frac{{dy}}{{dx}} \\

3x^2 + \frac{1}{{2\sqrt {2xy} }}(2xy' + 2y) + 2yy' = 0 \\

- 3x^2 - \frac{{2y}}{{2\sqrt {2xy} }} = \frac{{2xy'}}{{2\sqrt {2xy} }} + 2yy' \\

- 3x^2 - \frac{{2y}}{{2\sqrt {2xy} }} = y'\left( {\frac{{2x}}{{2\sqrt {2xy} }} + 2y} \right) \\

- 3x^2 - \frac{{2y}}{{2\sqrt {2xy} }} = y'\left( {\frac{{x + 2y\sqrt {2xy} }}{{\sqrt {2xy} }}} \right) \\

\frac{{ - 3x^2 (2\sqrt {2xy} ) - 2y}}{{2\sqrt {2xy} }} = y'\left( {\frac{{x + 2y\sqrt {2xy} }}{{\sqrt {2xy} }}} \right) \\

\left( {\frac{{ - 3x^2 (2\sqrt {2xy} ) - 2y}}{{2\sqrt {2xy} }}} \right)\left( {\frac{{\sqrt {2xy} }}{{x + 2y\sqrt {2xy} }}} \right) = y' \\

\left( {\frac{{ - 6x^2 y\sqrt {2xy} ) - 2xy^2 }}{{x\sqrt {xy} + 4y^2 x}}} \right) = y' \\

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

\end{array}\)

Thanks for your time, sophie

 

[mark07]

Derivatives are correct. Last two equations are wrong. Redo the multiplication of those two parenthesis terms right above them. Simplify numbers/expressions as much as possible before you multiply to ease your job.

 

[Sophie]

Thanks, Mark. Here is the fix


\(\displaystyle \L\\\begin{array}{l}
\left( {\frac{{ - 3x^2 \left( {2\sqrt {2xy} } \right) - 2y}}{{2\sqrt {2xy} }}} \right)\left( {\frac{{\sqrt {2xy} }}{{x + 2y\sqrt {2xy} }}} \right) = y' \\
\left( {\frac{{ - 3x^2 \left( {2\sqrt {2xy} } \right) - 2y}}{2}} \right)\left( {\frac{1}{{x + 2y\sqrt {2xy} }}} \right) = y' \\
\left( {\frac{{ - 3x^2 \left( {\sqrt {2xy} } \right) - y}}{1}} \right)\left( {\frac{1}{{x + 2y\sqrt {2xy} }}} \right) = y' \\
\left( {\frac{{ - 3x^2 \left( {\sqrt {2xy} } \right) - y}}{{x + 2y\sqrt {2xy} }}} \right) = y' \\
\end{array}\)

Thanks Sophie