To differentiate the function $y = (6x - 5)^2(3 - x^5)^2$, we will use the product rule and the chain rule.

Given:

$y = (6x - 5)^2(3 - x^5)^2$

Let's define two functions:

• $u = (6x - 5)^2$
• $v = (3 - x^5)^2$

The derivative of $y$ with respect to $x$ is found using the product rule: $\frac{dy}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}$

Step 1: Differentiate $u = (6x - 5)^2$

Using the chain rule: $\frac{du}{dx} = 2(6x - 5) \cdot \frac{d}{dx}(6x - 5)$ $\frac{d}{dx}(6x - 5) = 6$ $\frac{du}{dx} = 2(6x - 5) \cdot 6 = 12(6x - 5)$

Step 2: Differentiate $v = (3 - x^5)^2$

Using the chain rule: $\frac{dv}{dx} = 2(3 - x^5) \cdot \frac{d}{dx}(3 - x^5)$ $\frac{d}{dx}(3 - x^5) = -5x^4$ $\frac{dv}{dx} = 2(3 - x^5) \cdot (-5x^4) = -10x^4(3 - x^5)$

Step 3: Apply the product rule

Now, apply the product rule: $\frac{dy}{dx} = 12(6x - 5)(3 - x^5)^2 + (6x - 5)^2 \cdot (-10x^4)(3 - x^5)$

Simplify: $\frac{dy}{dx} = 12(6x - 5)(3 - x^5)^2 - 10x^4(6x - 5)^2(3 - x^5)$

This is the derivative of the given function.

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How does the product rule apply to other types of functions?
2. What is the chain rule, and how is it used in differentiation?
3. How can you simplify complex derivatives?
4. What are some practical applications of differentiation in physics?
5. How does implicit differentiation work?

Tip: Always simplify your derivative step by step to avoid mistakes when applying multiple rules like the product and chain rules.