Question Statement

Find $\frac{dy}{dx}$ for the implicitly defined curve:

$x+xy-y^2-20=0$

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Background and Explanation

This problem requires implicit differentiation, a technique used when $y$ is defined implicitly as a function of $x$ rather than explicitly as $y=f(x)$. Since $y$ depends on $x$, every time we differentiate a term containing $y$, we must apply the chain rule by multiplying by $\frac{dy}{dx}$.

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Solution

We start with the given equation and differentiate both sides with respect to $x$.

Step 1: Differentiate each term with respect to $x$

Given: $x+xy-y^2-20=0$

Differentiating term by term:

• $\frac{d}{dx}(x)=1$
• $\frac{d}{dx}(xy)$ requires the product rule: $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$
• $\frac{d}{dx}(y^2)$ requires the chain rule: $\frac{d}{dx}(y^2)=2y\cdot \frac{dy}{dx}$
• $\frac{d}{dx}(20)=0$ (derivative of a constant)

Applying these rules:

$\frac{d}{dx}(x)+\frac{d}{dx}(xy)-\frac{d}{dx}(y^2)-\frac{d}{dx}(20)=0$

Step 2: Apply product rule to the $xy$ term

For $\frac{d}{dx}(xy)$, let $u=x$ and $v=y$: $\frac{d}{dx}(xy)=x\cdot \frac{dy}{dx}+y\cdot \frac{dx}{dx}=x\frac{dy}{dx}+y$

Substituting back:

$1+\left(y\cdot \frac{d}{dx}(x)+x\cdot \frac{d}{dx}(y)\right)-2y^{2-1}\frac{dy}{dx}-0=0$

Step 3: Simplify the derivatives

Since $\frac{d}{dx}(x)=1$ and $\frac{d}{dx}(y)=\frac{dy}{dx}$:

$1+y\cdot 1+x\cdot \frac{dy}{dx}-2y\frac{dy}{dx}=0$

This simplifies to:

$1+y+x\frac{dy}{dx}-2y\frac{dy}{dx}=0$

Step 4: Group terms containing $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ from the terms that contain it:

$1+y+(x-2y)\frac{dy}{dx}=0$

Step 5: Isolate $\frac{dy}{dx}$

Move the non-derivative terms to the right side:

$(x-2y)\frac{dy}{dx}=-1-y$

Divide both sides by $(x-2y)$:

$\frac{dy}{dx}=\frac{-(1+y)}{x-2y}$

This can also be written as:

$\frac{dy}{dx}=\frac{1+y}{2y-x}$

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Key Formulas or Methods Used

• Implicit Differentiation: Differentiating both sides of an equation with respect to $x$, treating $y$ as a function of $x$
• Product Rule: $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ (applied to the $xy$ term)
• Chain Rule: $\frac{d}{dx}(f(y))=f^{\prime }(y)\cdot \frac{dy}{dx}$ (applied to the $y^2$ term, giving $2y\frac{dy}{dx}$)
• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$

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Summary of Steps

1. Differentiate both sides of the equation $x+xy-y^2-20=0$ with respect to $x$
2. Apply the product rule to $xy$ to get $y+x\frac{dy}{dx}$
3. Apply the chain rule to $y^2$ to get $2y\frac{dy}{dx}$
4. Collect all terms containing $\frac{dy}{dx}$ on one side and remaining terms on the other
5. Factor out $\frac{dy}{dx}$ and solve by dividing both sides by the coefficient of $\frac{dy}{dx}$