\(\bf x^2+2x6-y^2+x=20\impliedby taking\ \cfrac{dy}{dx}\textit{ to both sides}
\\\\\\ 2x+2\left( y+x\frac{dy}{dx}\right)-2y\frac{dy}{dx}=0\impliedby \textit{common factor 2}
\\\\\\
2\left[ x+\left( y+x\frac{dy}{dx}\right)-y\frac{dy}{dx}\right]=0
\\\\\\ x+y+x\frac{dy}{dx}-y\frac{dy}{dx}=0\impliedby \textit{common factor }\frac{dy}{dx}
\\\\\\
\cfrac{dy}{dx}(x-y)=-x-y\implies \cfrac{dy}{dx}=\cfrac{-x-y}{x-y}\qquad
\begin{cases}
x=3\\
y=4
\end{cases}\ then
\\\\\)
\(\bf \cfrac{dy}{dx}=\cfrac{-3-4}{3-4}\implies \cfrac{dy}{dx}=7\\\\
-----------------------------\\\\
\textit{tangent equation at }\begin{cases}
x=3\\
y=4\\
m=7
\end{cases}
\\\\\\
y-4=7(x-3)\implies y=7x-21+4\implies y=7x-17\)
