Finding the equation of a line that is tangent to.

Question

I have the following question & its answer but I do not understand how some parts were obtained - Q&A, Previous Q

Where does the $1+dy/dx (x-1)$ come from?

Answer 1

ok so you don't have any problem with calculating $\frac{dy}{dx}$

you want to find the tangent at the curve at the point (1,1)

the equation of a line passing through the point (1,1) is given by

$$(y-1)=m(x-1)$$ where m is the slope of the line

now to calculate$$ m=slope=\frac{dy}{dx}\big|_{(1,1)}$$ $$ m=slope=-\frac{2x+y}{x+y}\big|_{(1,1)}$$ $$ m=slope=-\frac{2(1)+(1)}{(1)+(1)}=\frac{-3}{2}$$

thus equation of tangent $$(y-1)=-\frac{3}{2}(x-1)$$ $$y-1=-\frac{3}{2}x+\frac{3}{2}$$ $$y=-\frac{3}{2}x+\frac{5}{2}$$

Answer 2

Hint: Assuming $y=y(x)$ then we get by the chain and product rule

$4x+2y(x)y'(x)+2y(x)+2xy'(x)=0$ Can you finish?