https://file.aiquickdraw.com/mathbot/file/17274513412970lujqv3o.jpg

The equation in the image is:

\[
z = \frac{x}{x + y}
\]

We need to compute the derivative of \( z \) with respect to \( x \), assuming \( y \) is a constant. This is a quotient rule derivative problem, as the function \( z \) is a ratio of two functions of \( x \).

The quotient rule for derivatives is given by:

\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]

Here, 
- \( u = x \) 
- \( v = x + y \)

Now, let's compute the derivative step by step:

\[
\frac{dz}{dx} = \frac{(x + y) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(x + y)}{(x + y)^2}
\]

Since \( \frac{d}{dx}(x) = 1 \) and \( \frac{d}{dx}(x + y) = 1 \) (as \( y \) is constant), the derivative simplifies to:

\[
\frac{dz}{dx} = \frac{(x + y) \cdot 1 - x \cdot 1}{(x + y)^2}
\]

Simplifying further:

\[
\frac{dz}{dx} = \frac{x + y - x}{(x + y)^2} = \frac{y}{(x + y)^2}
\]

Thus, the derivative of \( z \) with respect to \( x \) is:

\[
\frac{dz}{dx} = \frac{y}{(x + y)^2}
\]

Would you like more details on this, or do you have any specific questions?

Here are 5 related questions for further exploration:
1. How would the derivative change if \( y \) were also a function of \( x \)?
2. What happens to the derivative as \( y \to 0 \)?
3. Can you apply the chain rule to compute a second derivative of \( z \) with respect to \( x \)?
4. How does the quotient rule compare with the product rule in solving such derivatives?
5. Can you find the partial derivative of \( z \) with respect to \( y \)?

**Tip:** The quotient rule is essential when differentiating functions that are ratios of two differentiable functions. Always identify \( u \) and \( v \) clearly.

Solve the derivative of z = x / (x + y) with respect to x.

Learn how to solve the derivative of z = x / (x + y) using the quotient rule. This calculus problem covers the steps of applying the quotient rule to differentiate functions involving ratios of variables. Suitable for students in Grades 11-12.

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