Question

Calculus Questions using implicit differentiation to find the equation of the tangent line that is given curve at the 2, -1

Answer 1

The tangent line to the curve has slope equal to
`(\mathrm dy)/(\mathrm dx)` at the point (2, -1). We have, by implicit differentiation,

`3y^2-6xy=2x+11\implies6y(\mathrm dy)/(\mathrm dx)-6y-6x(\mathrm dy)/(\mathrm dx)=2`

`\implies(\mathrm dy)/(\mathrm dx)=(2+6y)/(6y-6x)=(1+3y)/(3y-3x)`

At the point (2, -1), the slope is then 2/9, so the tangent line has equation

`y-(-1)=\frac29(x-2)\implies y=\frac29x-\frac{13}9`