Part 1: Implicit Differentiation

We are given the equation:

$\frac{x^2}{x + y} = y^2 + 1$

To find the derivative of $y$ with respect to $x$ (i.e., $\frac{dy}{dx}$), we'll apply implicit differentiation step by step.

Step 1: Differentiate the left-hand side

$\frac{d}{dx} \left( \frac{x^2}{x + y} \right)$

This is a quotient, so we apply the quotient rule, where for $\frac{f(x)}{g(x)}$, the derivative is:

$\frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$

Let $f(x) = x^2$ and $g(x) = x + y$. We need the derivatives of both:

• $f'(x) = 2x$
• $g'(x) = 1 + \frac{dy}{dx}$ (since $y$ is a function of $x$)

Now apply the quotient rule:

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

Step 2: Differentiate the right-hand side

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

The derivative of $y^2 + 1$ is $2y \cdot \frac{dy}{dx}$ since $y$ is a function of $x$.

Step 3: Set up the equation

Now we can equate the derivatives of both sides:

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

Part 2: Evaluate at the point $(1, 0)$

Substitute $x = 1$ and $y = 0$ into the equation and solve for $\frac{dy}{dx}$.

Step 1: Substitute $x = 1$ and $y = 0$ into the equation

First, substitute $x = 1$ and $y = 0$ into the differentiated equation:

$\frac{(2(1))(1 + 0) - 1^2(1 + \frac{dy}{dx})}{(1 + 0)^2} = 2(0) \cdot \frac{dy}{dx}$

Simplifying:

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

This simplifies to:

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

$1 - \frac{dy}{dx} = 0$

$\frac{dy}{dx} = 1$

So, the derivative $\frac{dy}{dx}$ at the point $(1, 0)$ is $1$.

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Summary:

• Part 1: The implicit derivative is found using the quotient rule and implicit differentiation.
• Part 2: The derivative evaluated at the point $(1, 0)$ is $\frac{dy}{dx} = 1$.

Do you want more details or have any questions about the steps?

Related Questions:

1. How do you apply the product rule and chain rule in implicit differentiation?
2. Can you differentiate more complex implicit equations involving higher powers of $y$?
3. How would the process change if evaluated at a different point?
4. What happens if $y$ is explicitly defined in terms of $x$? Does the process differ?
5. Can you explain how implicit differentiation works for multivariable functions?

Tip:

Always remember to apply the chain rule when differentiating terms that involve $y$, as $y$ is a function of $x$.