Answer
a) $y'=\frac{-1-4x-y}{x}$
b) $y'=\frac{-2x^2-1}{x^2}$
c) See below.
Work Step by Step
a)
$\frac{d}{dx}(2x^2+x+xy=1)$
$4x+1+y+xy'=0$
If we solve for $y'$, we get $y'=\frac{-1-4x-y}{x}$
b)
First, we must solve for $y$. We get $y=\frac{1-2x^2-x}{x}$. We take the derivative of $y$ by using the Quotient Rule.
$y'=\frac{x(-4x-1)-(1-2x^2-x)}{x^2}$
$y'=\frac{-2x^2-1}{x^2}$
c) From part (a), we found that $y'=\frac{-1-4x-y}{x}$. We'll plug in $y=\frac{1-2x^2-x}{x}$ (which we found in part (b)) into this equation.
$y'=\frac{-1-4x-\frac{1-2x^2-x}{x}}{x}$
$y'=\frac{\frac{-x-4x^2-1+2x^2+x}{x}}{x}$
$y'={\frac{-x-4x^2-1+2x^2+x}{x^2}}=\frac{-2x^2-1}{x^2}$
Since this matches the answer we got in part (b), we see that our answers are consistent.
