Question

Find an equation tangent to the curve at the given point. 1. x²y+xy²=3x @ (-1,1)

Answer 1

Compute
`(\mathrm dy)/(\mathrm dx)` via implicit differentiation:

`x^2y+xy^2=3x\implies(\mathrm d)/(\mathrm dx)[x^2y+xy^2]=(\mathrm d)/(\mathrm dx)[3x]`

`(\mathrm d)/(\mathrm dx)[x^2y]+(\mathrm d)/(\mathrm dx)[xy^2]=3`

`\left((\mathrm d)/(\mathrm dx)[x^2]y+x^2(\mathrm d)/(\mathrm dx)[y]\right)+\left((\mathrm d)/(\mathrm dx)[x]y^2+x(\mathrm d)/(\mathrm dx)[y^2]\right)=3`

`\left(2xy+x^2(\mathrm dy)/(\mathrm dx)\right)+\left(y^2+2xy\,(\mathrm dy)/(\mathrm dx)\right)=3`

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

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

The value of
`(\mathrm dy)/(\mathrm dx)` at any point
`(x,y)` on the given curve is the slope of the line tangent to it at this point. So at (-1, 1), we get

`(\mathrm dy)/(\mathrm dx)=-4`

Then the equation of this tangent line is

`y-1=-4(x-(-1))\implies y=-4x-3`