Question

find the slope of the tangent to the graph of 3xy-2x+ 3y^2=5 at the point (2,1).
a) -1/3
b) -1/12
c) 1/12
d) 1
e) undefined

Answer 1

\(\bf 3xy-2x+ 3y^2=5\implies (3y^2+3xy+\square^2)=2x+5
\\ \quad \\
3(y^2+xy+\square^2)=2x+5\implies (y^2+xy+\square^2)=\cfrac{2x+5}{3}
\\ \quad \\
\textit{so completing the square we get}
\\ \quad \\
\left[y^2+xy+\left(\frac{x}{2}\right)^2\right]-\left(\frac{x}{2}\right)^2=\cfrac{2x+5}{3}
\\ \quad \\
\left[y^2+xy+\left(\frac{x}{2}\right)^2\right]=\cfrac{2x+5}{3}+\left(\cfrac{x}{2}\right)^2\implies \left(y+\frac{x}{2}\right)^2=\cfrac{2x+5}{3}+\cfrac{x^2}{4}
\\ \quad \\
y+\cfrac{x}{2}=\sqrt{\cfrac{2x+5}{3}+\cfrac{x^2}{4}}\implies y=\sqrt{\cfrac{2x+5}{3}+\cfrac{x^2}{4}}-\cfrac{x}{2}\)

so you'd get the derivative of "y" and then set x = 2 to get the slope at (2, 1)