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To solve the given problem, we need to differentiate \( xy = 5 \) implicitly with respect to \( x \).

### Step-by-step solution:

Given:
\[
xy = 5
\]

We apply implicit differentiation to both sides of the equation with respect to \( x \).

\[
\frac{d}{dx}(xy) = \frac{d}{dx}(5)
\]

The right-hand side is a constant, so its derivative is 0:
\[
\frac{d}{dx}(5) = 0
\]

Now, for the left-hand side, we use the **product rule** of differentiation because we are differentiating a product of \( x \) and \( y \).

The product rule states:
\[
\frac{d}{dx}(uv) = u'v + uv'
\]
where \( u = x \) and \( v = y \). Thus:
\[
\frac{d}{dx}(xy) = \frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y)
\]
This simplifies to:
\[
1 \cdot y + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx}
\]

So, we have:
\[
y + x \frac{dy}{dx} = 0
\]

### Solving for \( \frac{dy}{dx} \):

Now, isolate \( \frac{dy}{dx} \) by subtracting \( y \) from both sides:
\[
x \frac{dy}{dx} = -y
\]

Finally, divide by \( x \) to solve for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{-y}{x}
\]

Thus, the derivative \( \frac{dy}{dx} \) is:
\[
\frac{dy}{dx} = \frac{-y}{x}
\]

Would you like further clarification or details?

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Here are 5 related questions:
1. How would you find \( \frac{dy}{dx} \) if the equation was \( xy = 10 \)?
2. What is the product rule for differentiation, and how does it apply to implicit differentiation?
3. How would the process change if you differentiated \( x^2y = 6 \)?
4. Could you explain how to use implicit differentiation on trigonometric functions, such as \( \sin(xy) = 1 \)?
5. How can implicit differentiation be used to find the slope of a tangent line to a curve at a given point?

**Tip:** Implicit differentiation is useful when it is difficult or impossible to solve for one variable explicitly before differentiating.

Find dy/dx by implicit differentiation given that xy = 5.

Learn how to find the derivative dy/dx using implicit differentiation for the equation xy = 5. This solution involves the product rule and is ideal for high school calculus students.

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