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

[iceman99]

Find the derivative of y=(3x+4)^5/(2x^2-3x)^6



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?

 

[iceman99]

the -14x^2 in my final answer should be -24, but yeah that does not change much.

 

[Dick]

Your answer looks good to me. They took more than an x out of the bottom. (2x^2-3x)^7=x^7*(2x-3)^7. I think their answer is simply wrong.

 

[HallsofIvy]

5 hours between your first and third posts? Do you expect people to be sitting around waiting for you to post a problem?

Find the derivative of y=(3x+4)^5/(2x^2-3x)^6



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

Okay, that's a good idea. I have always preferred that to the quotient rule.

(3x+4)^5[-6(2x^2-3x)^-7(4x-3)]+(2x^2-3x)^-6[5(3x+4)^4(3)][\quote]
What does this have to do with the product rule? (fg)'= f'g+ fg'.
Here f= (3x+ 4)^5 so f'= (3)(5)(3x+4)^4= 15(3x+4)^4. f'g= 15(3x+4)^4(2x-3x)^(-6). g'= -6(2x^2- 3x)(4x-3) so fg'= 6(2x^2-3x)(4x-3)(3x+4)^5

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?

 

[iceman99]

No, I do not expect people to be waiting around for me to post a problem; I was merely annoyed that no one had replied in my five hours.

Rather than the quotient rule is all I meant by my phrase.

 

[iceman99]

In my g, I forgot the square so it should be (2x^2-3x)^-6
and consequently would make g’=-6(2x^2-3x)^-7(4x-3) correct?

 

[Dick]

I wasn't checking all of the steps, but "3(3x+4)^4 (-14x^2-29x+24) all over (2x^2-3x)^7" is what I got.

 

[iceman99]

Ok, thank you very much, I must have written the answer down incorrectly.

 

[Dick]

Ok, thank you very much, I must have written the answer down incorrectly.

I'm quoting what you wrote down.