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To fill in the blank in the given expression, let's solve the problem step by step.

We have:
1. \(\frac{d}{dx}(x^2) = 2x\)
2. \(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\)

The expression we need to differentiate is:
\[
\frac{d}{dx}\left((9x^2 + 5y^2 - x)^2\right)
\]
Using the chain rule, we differentiate it:
\[
\frac{d}{dx}\left((9x^2 + 5y^2 - x)^2\right) = 2(9x^2 + 5y^2 - x) \cdot \frac{d}{dx}(9x^2 + 5y^2 - x)
\]

Now, we differentiate the inner expression:
\[
\frac{d}{dx}(9x^2 + 5y^2 - x) = 18x + 10y \frac{dy}{dx} - 1
\]

Thus, the original derivative becomes:
\[
2(9x^2 + 5y^2 - x)(18x + 10y \frac{dy}{dx} - 1)
\]

To match the form given in the problem:
\[
2x + 2y \frac{dy}{dx} = 2(9x^2 + 5y^2 - x)\left(18x + (\text{BLANK}) \frac{dy}{dx} - 1\right)
\]

From the above derivation, we can see that the BLANK should be \(10y\).

So, the completed expression is:
\[
2x + 2y \frac{dy}{dx} = 2(9x^2 + 5y^2 - x)\left(18x + 10y \frac{dy}{dx} - 1\right)
\]

Would you like a more detailed explanation or have any questions?

Here are 5 related questions to consider:
1. How do you apply the chain rule for derivatives?
2. What is the derivative of a product of functions?
3. How can you differentiate implicit functions?
4. Why is the chain rule useful in calculus?
5. What happens when you differentiate higher powers, such as \((9x^2 + 5y^2 - x)^3\)?

**Tip:** Always simplify your derivative as much as possible before substituting values to avoid errors.

Fill in the blank in the following expression: 2x + 2y dy/dx = 2(9x^2 + 5y^2 - x)(18x + (BLANK) dy/dx - 1).

Learn how to apply implicit differentiation and the chain rule to solve a complex expression involving x and y. Fill in the missing term and simplify the derivative. Suitable for students in Grades 10-12.

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