(3x-4) (x'2-6x+7)

use x^2 to show "x squared" etc

I would expand it first, just like in my previous reply , then take the derivative.
check your answer by doing it both ways, like I showed you

9x'2-44x+45

To multiply the expression (3x - 4) and (x^2 - 6x + 7), we can use the distributive property.

We can consider each term in the first expression, (3x - 4), and multiply it by each term in the second expression, (x^2 - 6x + 7).

So, let's start:

First, multiply the term 3x from the first expression by each term in the second expression:
3x * x^2 = 3x^3
3x * -6x = -18x^2
3x * 7 = 21x

Next, multiply the term -4 from the first expression by each term in the second expression:
-4 * x^2 = -4x^2
-4 * -6x = 24x
-4 * 7 = -28

Now, we add up all the terms we obtained:
3x^3 + (-18x^2) + 21x + (-4x^2) + 24x + (-28)

Simplifying this expression, we get:
3x^3 - 22x^2 + 45x - 28

Therefore, the result of multiplying (3x - 4) and (x^2 - 6x + 7) is 3x^3 - 22x^2 + 45x - 28.