Find the derivative of

y=cot³(1-2x)²
y'=3[cot(1-2x)²]² (-csc²(1-2x)²)(2-4x)(-2)
y'=(-6+24x)[cot(1-2x)²]²(-csc²(1-2x)²)

How do you simplify further?

I believe your last step contains and error. Shouldn't the first parentheses contain (-12 +24x) ?

The "2" exponent outside the first bracket could have been written "cot^2"

I don't see any furher simplifications

Thanks. The answer key says the 12 is positive.

To simplify the expression further, you can use the properties of trigonometric functions and exponent rules. Let's break down the expression step by step:

First, let's rewrite the expression:

y' = (-6 + 24x)[cot(1-2x)²]²(-csc²(1-2x)²)

Now, let's simplify each part of the expression:

1. Simplifying [cot(1-2x)²]²:
- Remember that cot(x) = 1/tan(x), and cot²(x) = 1/tan²(x).
- So, [cot(1-2x)²]² = [1/tan(1-2x)²]²
- Applying the rule (1/x)² = 1/x², we get:
[1/tan(1-2x)²]² = 1/[tan(1-2x)²] = 1/(tan²(1-2x))

2. Simplifying (-csc²(1-2x)²):
- Since csc(x) = 1/sin(x) and csc²(x) = 1/sin²(x), we have:
(-csc²(1-2x)²) = -1/sin²(1-2x)².

3. Putting it all together:
- Now we can substitute the simplified versions back into the original expression:
y' = (-6 + 24x) * (1/[tan²(1-2x)]) * (-1/sin²(1-2x)²)

The expression y' = (-6 + 24x) * (1/[tan²(1-2x)]) * (-1/sin²(1-2x)²) is the simplified form of the derivative.