I want to find the slop of the tangent line to the curve $2x^{3}y+4.5y-xy^{2}=8 $ at point (0,16/5) using Barrow's method. Here's what I've done.
Substitute x with x+e and y with y+a $2(x^{3}+3ex^{2}+3e^{2}+e^{3})(y+a)+4.5(y+a)-(x+e)(y^{2}+2ay+y^{2})=8$ $2(x^{3}y+x^{2}a+3ex^{2}y+3ex^{2}a+3e^{2}y+3e^{2}a+e^{3}y+e^{3}a)+4.5y+4.5a-y^{2}x-2ayx-y^{2}x-ey^{2}-2aye-ey^{2}=8$
I now remove terms with powers of e or a or products of the two $2(x^{3}y+x^{2}a+3ex^{2}y+3ex^{2}a)+4.5y+4.5a-y^{2}x-2ayx-y^{2}x-ey^{2}-2aye-ey^{2}=8$
Remember $2x^{3}y+4.5y-xy^{2}=8$ so I can replace those terms with 8 which sets the equation equal to 0 when subtracted $2x^{2}a+6ex^{2}y+6ex^{2}a+4.5a-y^{2}x-2ayx-ey^{2}-2aye-ey^{2}=8$
I'm confused as to where I go from here
Barrow sets $f(x+e,y+a)=f(x,y)$ and discards terms that are quadratic or higher in $e$ and $a$. This equation is equivalent to $f(x+e,y+a)-f(x,y)=0$, so subtract the original function from what you’ve ended up with. In the resulting expression, $x$ and $y$ are the coordinates of the point at which you’re computing the tangent, while $e$ and $a$ are “infinitesimal” displacements from it. Once you plug in the values of $x$ and $y$ for the point, you should end up with an equation of a line in $e$ and $a$. I hope that you know how to extract the slope from that yourself.