expanding expressions: 3(x + h)^3 + 2(x + h) - ....

[gopher]

We are working on calculating derivatives, and I am getting stuck on the algebra part. Specifically, I have:

\(\displaystyle 3(x\, +\, h)^3\, +\, 2(x\, +\, h)\, -\, 1\, -\, (3x^3\, +\, 2x\, -\, 1)\)

...and I need to simplify this to:

\(\displaystyle 3(x^3\, +\, 3x^2h\, +\, 3xh^2\, +\, h^3)\, +\, (2x\, +\, 2h)\, -\, 1\, -\, 3x\, -\, 2x\, +\, 1\)

I see how "2(x + h)" becomes "(2x+2h)". I'm told that "3(x + h)^3" can be figured out with the help of Pascal's Triangle. Does that sound right?

But why does "-(3x^3 + 2x - 1)" turn into "-3x^3 - 2x + 1"? When we take out parentheses, it looks like we change the sign of the 2x and 1. Why not the 3x^3?
Did I skip a class here?

Thank you!

 

[arthur ohlsten]

eq1) y=3x^3 +2x -1
replace y with y+dy , and x with x+h
eq2) y+dy =3[x+h]^3+2[x+h]-1

subtract eq1 from eq2
dy = 3[x+h]^3+2[x+h]-1 -[3x^3+2x-1]

first let us determine [x+h]^3
[x+h]^3=x^3+3x^2h+3xh^2+h^3
by pascals triangle
............1
..........1.....1
........1....2.....1
......1....3....3.....1

dy=3[x^3+3x^2h+3xh^2+h^3]+2[x+h]-1-[3x^3+2x-1] remove brackets
dy=3x^3+9x^2h+9xh^2+3h^3+2x+2h -1-3x^3-2x+1 combine likes
dy=9x^2h+9xh^2+3h^3+2h divide both sides by h
dy/h=9x^2+9xh+3h^2+2 let h approach 0
dy/dx=9x^2+9x[0] +3[0]^2+2
dy/dx=9x^2+2 answer

Arthur

 

[Mrspi]

Maybe this will help you see WHY the signs of each term change....

-(3x<SUP>2</SUP> + 2x - 1) means

-1*(3x<SUP>2</SUP> + 2x - 1)

Multiply each term inside the parentheses by -1:

-1(3x<SUP>2</SUP> + (-1)*(2x) - (-1)*1

-3x<SUP>2</SUP> + (-2x) - (-1)

-3x<SUP>2</SUP> - 2x + 1

In the original expression which was inside the parentheses, you had 3x<SUP>2</SUP>.....after removing the parentheses, you have -3x<SUP>2</SUP>. The sign of EACH TERM inside the parentheses got changed....