Find the derivative of y=(3x+4)^5/(2x^2-3x)^6(adsbygoogle = window.adsbygoogle || []).push({});

Some how the answer is simplified to 3x(6x^2-59x+24)/(2x-3)^7, I don’t know if I just copied it down wrong or I’m doing it incorrectly. Some assistance would be nice.

I changed the equation to y=(3x+4)^5*(2x-3x)^-6 and used the product rule/ chain rule

(3x+4)^5[-6(2x^2-3x)^-7(4x-3)]+(2x^2-3x)^-6[5(3x+4)^4(3)]

This simplified into

(-24x+18)(3x+4)^5divided by(2x^2-3x)^7 + 15(3x+4)^4 divided by (2x^2-3x)^6

Multiplying 15(3x+4)^4 divided by (2x^2-3x)^6 by (2x^2-3x) to the bottom and top get a common denominator

Which made it:

[(-24x+18)(3x+4)^5]+[(3x+4)^4(30x^2-45x)] all over (2x^2-3x)^7

At which point I took out the (3x+4)^4 which made it to:

((3x+4)^4) [(3x+4)(-24x+18)+(30x^2-45x] all over (2x^2-3x)^7

After foiling and simplifying the inside of the brackets I got

(3x+4)^4) (-72x^2-87x+72) taking out the three my final answer being:

3(3x+4)^4 (-14x^2-29x+24) all over (2x^2-3x)^7

I understand that in the answer they must have taken out an x at the bottom

Sorry if my steps are a little confusing. Did I do it right or did I miss an important piece?

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# High School calculus problem, basic for some, need a little help

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