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.