Use implicit differentiation to determine $\frac { d y } { d x }$ where $x ^ { 2 } y ^ { 3 } + x = y$ . Is $\frac { d y } { d x } = \frac { 2 x y ^ { 3 } + 1 } { - 3 x ^ { 2 } y ^ { 2 } + 1 }$ correct?

Question

Use implicit differentiation to determine

\frac { d y } { d x }

where

x ^ { 2 } y ^ { 3 } + x = y

. Is

\frac { d y } { d x } = \frac { 2 x y ^ { 3 } + 1 } { - 3 x ^ { 2 } y ^ { 2 } + 1 }

correct?

Quizplus · Accepted Answer

Differentiating both sides of the equation $x^2y^3 + x = y$ with respect to $x$ using the product rule and chain rule gives: $2xy^3 + x^2(3y^2)\frac{dy}{dx} + 1 = \frac{dy}{dx}$. Solving for $\frac{dy}{dx}$ indeed yields $\frac{dy}{dx} = \frac{2xy^3 + 1}{1 - 3x^2y^2}$, which matches the given expression when considering the negative sign in the denominator correctly.

Use Implicit Differentiation to Determine dydx\frac { d y } { d x }dxdy​

Use Implicit Differentiation to Determine $\frac { d y } { d x }$