x^2/(x+y) = y^2+1
Part 1: Derivative Calculate the derivative by implicit differentiation:
Part 2 : Evaluate the derivative at the point(1,0).

### 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 \(1\).

---

### 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 \(x\).

x^2/(x+y) = y^2+1. Part 1: Derivative Calculate the derivative by implicit differentiation. Part 2: Evaluate the derivative at the point (1,0).

Learn how to use implicit differentiation and the quotient rule to find the derivative of the equation x^2/(x+y) = y^2 + 1. This solution includes step-by-step differentiation and evaluates the derivative at the point (1, 0), making it ideal for students in Grades 11-12 studying calculus.

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