+118687 Firstly you do not need to be given y(5)=-26 You can work that out for yourself.
Like this, sub in x=5 and solve for y
\(5*5^2+5*5+5y=20\\ \text{divide through by 5}\\ 25+5+y=4\\ y=-26\)
So giving you that was probably a bit of a red herring.
\(5x^2+5x+xy=20\\ \text{differentiate both sides}\\ 10x+5+x\frac{dy}{dx}+1y=0\\ \text{Sub in x=5}\\ 50+5+5\frac{dy}{dx}+y=0\\ 5\frac{dy}{dx}=-55-y\\ \text{but we already know that when x=5, y=-26 so sub that in too}\\ 5\frac{dy}{dx}=-55+26\\ 5\frac{dy}{dx}=-29\\ \frac{dy}{dx}=-5.8\\ \)
y'(5) = -5.8
LaTex:
5x^2+5x+xy=20\\
\text{differentiate both sides}\\
10x+5+x\frac{dy}{dx}+1y=0\\
\text{Sub in x=5}\\
50+5+5\frac{dy}{dx}+y=0\\
5\frac{dy}{dx}=-55-y\\
\text{but we already know that when x=5, y=-26 so sub that in too}\\
5\frac{dy}{dx}=-55+26\\
5\frac{dy}{dx}=-29\\
\frac{dy}{dx}=-5.8\\
