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We are asked to find \(\frac{dy}{dx}\) given that:

\[
y = 4u^2 - 6u - 6 \quad \text{and} \quad u = 2x - 5.
\]

### Step 1: Use the Chain Rule

We will apply the chain rule since \(y\) is expressed as a function of \(u\), and \(u\) is a function of \(x\). The chain rule states:

\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.
\]

### Step 2: Find \(\frac{dy}{du}\)

Differentiate \(y = 4u^2 - 6u - 6\) with respect to \(u\):

\[
\frac{dy}{du} = \frac{d}{du}(4u^2) - \frac{d}{du}(6u) - \frac{d}{du}(6).
\]

Now, compute each derivative:
- \(\frac{d}{du}(4u^2) = 8u\),
- \(\frac{d}{du}(6u) = 6\),
- \(\frac{d}{du}(6) = 0\).

Thus:

\[
\frac{dy}{du} = 8u - 6.
\]

### Step 3: Find \(\frac{du}{dx}\)

Now, differentiate \(u = 2x - 5\) with respect to \(x\):

\[
\frac{du}{dx} = 2.
\]

### Step 4: Apply the Chain Rule

Using the chain rule:

\[
\frac{dy}{dx} = (8u - 6) \cdot 2.
\]

Substitute \(u = 2x - 5\) into this expression:

\[
\frac{dy}{dx} = (8(2x - 5) - 6) \cdot 2.
\]

Simplify:

\[
\frac{dy}{dx} = (16x - 40 - 6) \cdot 2 = (16x - 46) \cdot 2 = 32x - 92.
\]

Thus, the final result is:

\[
\frac{dy}{dx} = 32x - 92.
\]

Would you like further explanation on any part of this process?

Here are some related questions for further practice:

1. How would you find \(\frac{dy}{dx}\) if \(y = 5u^3 - 7u\) and \(u = 3x + 2\)?
2. What happens if the function for \(u\) is more complex, like \(u = \sin(x)\)?
3. How do we handle the chain rule when there are multiple variables involved?
4. Can you apply the product or quotient rule in conjunction with the chain rule?
5. What is the geometric interpretation of the derivative in this context?

**Tip**: Always simplify the inner function's derivative before applying it to the outer function to avoid mistakes.

Find dy/dx, if y = 4u^2 - 6u - 6 and u = 2x - 5.

This problem demonstrates how to apply the chain rule to differentiate the function y = 4u^2 - 6u - 6, where u is a linear function of x (u = 2x - 5). A step-by-step solution shows how to calculate dy/dx using differentiation techniques.

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