Originally Posted by
BinaryBoy Hi everyone,
I've been working with the chain and quotient rules mingled together in order to solve a fairly standard equation. I've managed to solve it, except for one part which involves the bottom of the equation which is normally squared, however, somehow it is cubed instead.
Following is my full working and my answer, along with the answer the text book showed me:
y = (x - 3)^3 / (2x + 1)^2
Inside - Outside / bottom ^ 2
u = (x + 3)^3
u` = 3(x - 3)^2
v = (2x + 1)^2
v` = 4(2x + 1)
(3(x - 3)^2 * (2x + 1)^2) - (4(2x + 1) * (x - 3)^3) / (2x + 1)^2
The next part involves factorizing. I look at it in terms of x and y to make the bulky equation easier to manage. Thus
x = (x - 3)
y = (2x + 1)
Thus:
3x^2y^2 - 4yx^3 / y^2
Simplify: find HCF...
x^2y (3y - 4x) / y^2
Return to normal:
(x - 3)^2 (2x + 1) [ 3(2x + 1) - 4(x - 3)] / (2x + 1)^2
(x - 3)^2 (2x + 1) [6x + 3 -4x + 12]
= (x - 3)^2 (2x + 1) (2x + 15) / (2x + 1)^2
However, the textbook's answer is:
(x - 3)^2 (2x + 15) / (2x + 1)^3
What I don't understand is how the (2x + 1) on the top came to be transferred to the bottom to make the bottom value (2x + 1) ^ 3 instead of the traditional ^2.
Any help on this will be much appreciated!
Many thanks,
Nathaniel