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We are given the equation \( xy = 1 \), and we are asked to find the value of \( y' + \frac{dy}{dx} \).

First, let's implicitly differentiate both sides of the equation \( xy = 1 \) with respect to \( x \).

### Step 1: Differentiate both sides
Using the product rule for differentiation:
\[
\frac{d}{dx}(xy) = \frac{d}{dx}(1)
\]

The product rule says \( \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \frac{dx}{dx} = x \frac{dy}{dx} + y \), and the derivative of the constant \( 1 \) is zero:
\[
x \frac{dy}{dx} + y = 0
\]

### Step 2: Solve for \( \frac{dy}{dx} \)
We can rearrange this equation to solve for \( \frac{dy}{dx} \):
\[
x \frac{dy}{dx} = -y
\]
\[
\frac{dy}{dx} = -\frac{y}{x}
\]

### Step 3: Substitute into the original expression
We are asked to find \( y' + \frac{dy}{dx} \), which is the same as \( \frac{dy}{dx} + \frac{dy}{dx} \). From the differentiation, we know \( y' = \frac{dy}{dx} \), so:
\[
y' + \frac{dy}{dx} = \frac{dy}{dx} + \frac{dy}{dx} = 2 \cdot \frac{dy}{dx}
\]

Now, substitute \( \frac{dy}{dx} = -\frac{y}{x} \) into this expression:
\[
2 \cdot \frac{dy}{dx} = 2 \cdot \left(-\frac{y}{x}\right) = -\frac{2y}{x}
\]

### Step 4: Conclusion
From the original equation \( xy = 1 \), we know that \( y = \frac{1}{x} \). Substituting this into the expression:
\[
-\frac{2y}{x} = -\frac{2 \cdot \frac{1}{x}}{x} = -\frac{2}{x^2}
\]

Thus, the expression for \( y' + \frac{dy}{dx} \) simplifies to \( -1 \).

Therefore, the answer is:
\[
\boxed{-1}
\]

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

Here are some related questions to think about:
1. How do you apply the product rule in implicit differentiation?
2. What other ways could the equation \( xy = 1 \) be solved using derivatives?
3. What happens to the derivative when different functions of \( x \) and \( y \) are multiplied?
4. Can implicit differentiation be applied to more complex equations like trigonometric functions?
5. What role does the constant term play in implicit differentiation?

**Tip**: Always verify the basic identity or equation provided before applying differentiation techniques.

Learn how to solve the problem 'If xy = 1, then y' + dy/dx is equal to'. This problem involves implicit differentiation using the product rule and solving for y' in terms of x and y. Suitable for calculus students in Grades 11-12 or college.

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