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77julia77 [94]
2 years ago
6

If y=(x^2-4)^5(3x+4)^4

Mathematics
1 answer:
alexira [117]2 years ago
5 0

Answer:

y'=2(3x+4)^3(x^2-4)^4(21x^2+20x-24)

Step-by-step explanation:

y=(x^2-4)^5(3x+4)^4

y'=[\frac{d}{dx}(x^2-4)^5](3x+4)^4+(x^2-4)^5[\frac{d}{dx}(3x+4)^4]

y'=10x(x^2-4)^4(3x+4)^4+(x^2-4)^5(12(3x+4)^3)

y'=10x(x^2-4)^4(3x+4)^4+12(x^2-4)^5(3x+4)^3

y'=2(3x+4)^3(x^2-4)^4(21x^2+20x-24)

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olga nikolaevna [1]
Hey there!

Let's set up our expression:

(7a-6b+7)-(8a-2)

In order to simplify, we can use that subtraction sign and distribute it, using the distributive property. We have:

7a-6b+7-8a+2

Notice how it's plus two, because a negative times a negative two is a positive two. Now, it's a matter of finding the like terms and adding or subtracting them. These like terms can either have no variable, or have different coefficients but the same variable. That means our like terms are the 7a and -8a, and the 7 and 2. There's no like term for the 6b. That means we have:

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