Solve the equation explicitly for y and differentiate to get y' in terms of x. 2x2+x+xy=1

2x2+x+xy=1 Differentiate throughout with respect to x d(2x2)dx+dxdx+d(xy)dx=d(1)dx 4x+1+[y⋅dxdx+x⋅dydx]=0 4x+1+[y⋅1+x⋅dydx]=0 4x+1+y+x⋅dydx=0 x⋅dydx=−(4x+1+y) dydx=−(4x+1+y)}{x} dydx=−(4x+1+y)x

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parthe0o

Answered question

2021-11-09

Solve the equation explicitly for y and differentiate to get y' in terms of x.

2 x^{2} + x + x y = 1

Answer & Explanation

John Twitchell

Beginner2021-11-10Added 19 answers

2 x^{2} + x + x y = 1

Differentiate throughout with respect to x

\frac{d (2 x^{2})}{d x} + \frac{d x}{d x} + \frac{d (x y)}{d x} = \frac{d (1)}{d x}

4 x + 1 + [y \cdot \frac{d x}{d x} + x \cdot \frac{d y}{d x}] = 0

4 x + 1 + [y \cdot 1 + x \cdot \frac{d y}{d x}] = 0

4 x + 1 + y + x \cdot \frac{d y}{d x} = 0

x \cdot \frac{d y}{d x} = - (4 x + 1 + y)

\frac{d y}{d x} = \frac{- (4 x + 1 + y)}{}} {x}

\frac{d y}{d x} = \frac{- (4 x + 1 + y)}{x}

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