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y=x^3-3x^2+2x+5

2007-05-03 01:59:41 · 8 answers · asked by Anonymous in Science & Mathematics Mathematics

8 answers

it is;

dy/dx = 3x² - 6x + 2

.

2007-05-03 02:02:32 · answer #1 · answered by SB 3 · 0 0

y'=3x^2-6x+2

To differentiate a simple function like this, simply pull the power off of each x and multiply it by the coefficient. Then subtract one from the original exponent to form your new exponent. For constants the derivative is zero and for first order terms the derivative is simply the coefficient of the term.

2007-05-03 02:08:50 · answer #2 · answered by Pius Thicknesse 4 · 0 0

No need for long stories:-
dy/dx = 3x² - 6x + 2

2007-05-03 19:31:19 · answer #3 · answered by Como 7 · 0 0

y' = 3x^2 - 6x + 2

2007-05-03 02:02:24 · answer #4 · answered by metalluka 3 · 0 0

dy/dx = 3x² - 6x + 2

2007-05-03 02:05:37 · answer #5 · answered by Dave 6 · 0 0

y' =3x^2 -6X +2

2007-05-03 02:04:10 · answer #6 · answered by bignose68 4 · 0 0

y = x^3 -3x^2 + 2x + 5
y+dy = (x+dx)^3 - 3.(x+dx)^2 +2.(x+dx) + 5 NB should be using Greek delta at this stage
y+dy = x^3 + 3x^2dx+3dx^2+dx^3 -3x^2-6x.dx-3dx^2+2x+2dx +5
y = x^3 -3x^2+2x+5
dy = 3x^2.dx+3dx^2+dx^3-6x.dx-3dx^2x+2dx

dy/dx = 3x^2+3dx+dx^2-6x-3dx + 2
dx -> 0 dy/dx (using deltas) -> dy/dx = 3x^2-6x+2

2007-05-03 03:34:49 · answer #7 · answered by welcome news 6 · 0 1

This is a worrying question because it is a very basic one.

Is there a bigger problem here?

2007-05-03 03:08:00 · answer #8 · answered by fred 5 · 0 1

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