Use implicit differentiation to find y' where x^2 +6xy + 4y^3 = 20. Evaluate this at (x, y) =...

The value after differentiating at (x,y)=(2,1) is {eq}\frac{{ - 5}}{{12}}. {/eq}

Given: {eq}{x^2} + 6xy + 4{y^3} = 20 {/eq}

Differentiate it w.r.t. x,

{eq}\begin{align*} 2x\frac{{dx}}{{dx}} + 6x\frac{{dy}}{{dx}} + 6y\frac{{dx}}{{dx}} + 4\left( {3{y^2}} \right)\frac{{dy}}{{dx}} &= 0\\ 2x + 6xy' + 6y + 12{y^2}y' &= 0\\ 2\left( {x + 3xy' + 3y + 6{y^2}y'} \right) &= 0\\ x + 3xy' + 3y + 6{y^2}y' &= 0\\ x + 3y + y'\left( {3x + 6{y^2}} \right) &= 0\\ y'\left( {3x + 6{y^2}} \right) &= - \left( {x + 3y} \right)\\ y' &= \frac{{ - \left( {x + 3y} \right)}}{{3x + 6{y^2}}} \end{align*} {/eq}

The value of y' at (x,y)=(2,1) is:

{eq}\begin{align*} y' &= \frac{{ - \left( {2 + 3\left( 1 \right)} \right)}}{{3\left( 2 \right) + 6{{\left( 1 \right)}^2}}}\\ y' &= \frac{{ - \left( {2 + 3} \right)}}{{6 + 6}}\\ y' &= \frac{{ - 5}}{{12}} \end{align*} {/eq}