Help last q maths ass! (1 Viewer)

[Cattle]

hey everyone i know i have already asked this and well yeh i still dont understand so if any one can explain how to do part b it would be greatly appreciated. the ass is due tomorrow

i can do part a) but i dont get the answer that is used in b) 2) a) differentiate 2x^2 +1(all over) 3x^2- 4
b) Hence evaluate integration from 0to 1 (cant figure out how to do the sign thingy on here) x (over) (3x^2- 4)^2

 

[insert-username]

I don't get the answer in part b either:


2) a) differentiate 2x^2 +1(all over) 3x^2- 4
b) Hence evaluate integration from 0to 1 (cant figure out how to do the sign thingy on here) x (over) (3x^2- 4)^2


d/dx(2x² + 1)/(3x² - 4)

If y = u/v, then dy/dx = (vu' - uv')/v²

= [4x(3x² - 4) - 6x(2x² + 1)]/(3x² - 4)²

= [12x³ - 16x - 12x³ - 6x]/(3x² - 4)²

= -22x/(3x² - 4)²

Differentiating using the product rule gives the same result:

d/dx(2x² + 1)(3x² - 4)^-1

If y = uv, dy/dx = vu' + uv'

= -6x(2x² + 1)(3x² - 4)^-2 + 4x(3x² - 4)^-1

= (3x² - 4)^-2[-12x³ - 6x + 12x³ - 16x]

= -22x/(3x² - 4)²

I think part b is wrong. Can someone else try the differentiation, in case I've stuffed it up somewhere?


I_F

 

[Riviet]

I just checked the differentiation and it's definitely correct. Part b is just... weird. Probably wrong as insert-username said. Check the question, chooke.

 

[Riviet]

Oh wait... i just figured it out!

x/(3x²-4)² = -1/22[-22x/(3x²-4)²]

Now we take the integral from 0 to 1 of x/(3x²-4)² by substituting the derivadand (the thing we were differentiating in part [a] lol, just made up that word from integrand) into -1/22[-22x/(3x²-4)²]!!

 

[insert-username]

Yuck. Yuck yuck yuck yuck yuck. But smart thinking.


I_F

 

[Riviet]

It's also known as "algebraic manipulation"