Question

Find \(d y / d x\) by implicit differentiation. $$ x^{3}-x y+y^{2}=4 $$

Short answer

\( \frac{dy}{dx} = \frac{y-3x^{2}}{x-2y}\)

Step-by-step solution

Differentiate term by term

Each term of the given equation ought to be differentiated, bearing in mind the rules of differentiation. The derivative of \(x^{3}\) is \(3x^{2}\), the derivative of \(xy\) is \(y + x \frac{dy}{dx}\) using the product rule of differentiation, and the derivative of \(y^{2}\) is \(2y \frac{dy}{dx}\) by the chain rule. We have \(3x^{2}-y-x \frac{dy}{dx}+2y\frac{dy}{dx}=0\).

Rearrange to isolate \(dy/dx\)

Next, isolate \(dy/dx\) on one side to find its value. This is done by subtracting 3\(x^{2}\) and \(y\) from both sides and regrouping like terms. This gives \( \frac{dy}{dx} (x - 2y) = y - 3x^{2}\).

Solve for \(dy/dx\)

Finally, divide through by \(x-2y\), to find the value of \(dy/dx\). This gives \( \frac{dy}{dx} = \frac{y-3x^{2}}{x-2y}\)