# Help with chain and quotient rule problem - unsure of one part

• Nov 11th 2010, 12:05 AM
BinaryBoy
Help with chain and quotient rule problem - unsure of one part
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

(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

(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
• Nov 11th 2010, 12:14 AM
Prove It
Quote:

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

(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

(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

Don't use the same letter as your dummy variable, at least make it a capital so that it can be read...

For starters, the $\displaystyle v^2$ in the denominator should be $\displaystyle (2x + 1)^4$, not $\displaystyle (2x + 1)^2$.

The rest should follow.
• Nov 11th 2010, 12:24 AM
BinaryBoy
Sorry about that, Prove It. I'll keep it in mind.

Why is the denominator $(2x + 1)^4$? I'm running through the method and I don't see where the value is changed to the power of 4. Please clarify.
• Nov 11th 2010, 12:25 AM
Opalg
Quote:

Originally Posted by BinaryBoy
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 That's where the mistake comes. You have forgotten to square the denominator (which was already a square). So you should have (2x + 1)^4.

..
• Nov 11th 2010, 12:36 AM
BinaryBoy
Oh, I see! Thanks, Opalg and Prove It! So once I have done that, I simply subtract the top (2x +1)^1 value and the bottom (2x +1)^4 value. And 1 - 4 equals 3. Please clarify if that is incorrect reasoning.

Cheers!

Nathaniel
• Nov 11th 2010, 12:37 AM
Prove It
Well, actually 4 - 1 = 3, but your reasoning is correct.