Hi all, I will post the question then my method, I am going slightly wrong somewhere, but cant quite see where. Here goes
Find the gradient function (differentiate once)
[(x+3)(x-2)]/2x
so here is my method
(x^2 +3x - 2x - 6)/2x
= x/2 + 3/2 - 1 - 3x^-1
= 1/2(x) + 1/2 -3x^-1
dy/dx = 1/2(x^0) + 1/2 + 3x^-2
= 1/2(1) + 1/2 + 3/x^2
= 1 + 3/x^2
The answer on the back is 1/2 + 3/x^2
Im doing something wrong somewhere, but cant see where!
2007-05-11 08:00:36 · 10 answers · asked by John W 2 in Science & Mathematics ➔ Mathematics
im not sure why someone has plagued my question and given everyone a thumbs down. (I believe it is a dirty tactic used by the 6th answerer to get the best answer). I.e. If a question does not have a best answer chosen, the best answer is automatically selected as (number of thumbs up - thumbs down). I also believe you loose points for a thumb down. The 6th answer will be flagged
2007-05-11 08:18:29 · update #1
sorry i mean 8th answerer
2007-05-11 08:19:17 · update #2
On the line where you have:
dy/dx=1/2(x^0) + 1/2 + 3x^-2
The derivative of the middle term, 1/2, should be zero since the derivative of any constant is always zero.
So, just take out the 1/2 in the middle and your answer will be correct.
2007-05-11 08:03:52 · answer #1 · answered by victoria 5 · 1⤊ 1⤋
The derivative of a constant is zero, so from your fourth step:
dy/dx = 1/2 (x^0) + 0 + 3/x^2 (note the middle term)
= 1/2 (1) + 0 + 3/x ^2
= 1/2 + 3/x^2
2007-05-11 08:12:25 · answer #2 · answered by donxfive 2 · 0⤊ 1⤋
The derivative of a constant is 0. So 1/2 +1/2 does not happen. You simply have the 1/2 from differentiating (1/2)x.
2007-05-11 08:13:12 · answer #3 · answered by Kevin M 3 · 0⤊ 1⤋
3 lines from the end
the derivative of a 1/2 is 0 not 1/2
dy/dx = 1/2(x^0) + 3x^-2
2007-05-11 08:06:24 · answer #4 · answered by fred 5 · 1⤊ 1⤋
the derivative of 1/2 is zero, yo! it should go like this:
dy/dx = 1/2(x^0) + 3x^-2
= 1/2(1) + 3/x^2
= 1/2 + 3/x^2
2007-05-11 08:08:13 · answer #5 · answered by Anonymous · 2⤊ 1⤋
here is the place you make an error:
= 1/2(x) + 1/2 -3x^-1
dy/dx = 1/2(x^0) + 1/2 + 3x^-2
Here 1/2 must be zero. Because it is a number without x
it should be
dy/dx = 1/2(x^0) + 0+ 3x^-2
2007-05-11 08:06:40 · answer #6 · answered by iyiogrenci 6 · 1⤊ 1⤋
Their answer is correct.
(x^2 +3x - 2x - 6)/2x = (x² + x - 6) / 2x =
(x²/2x + x/2x - 6/2x) =
x/2 + 1/2 - 3/x =
(1/2)x + 1/2 - 3(x^-1)
Now differentiate:
dy/dx = 1/2 (1) + 0 - 3(-1)(x^-2)
dy/dx = 1/2 + 3x^-2
dy/dx = 1/2 + 3/x^2
So, their answer appears to be correct.
2007-05-11 08:32:39 · answer #7 · answered by MathBioMajor 7 · 0⤊ 0⤋
The derivative of 1/2 is 0...and you left it as 1/2...that's your problem...besides that you did good.
Take out the 1/2 and you have the right answer.
2007-05-11 08:06:08 · answer #8 · answered by Anonymous · 1⤊ 1⤋
This line is not correct.
y = 1/2(x) + 1/2 -3x^-1
dy/dx = 1/2(x^0) + 1/2 + 3x^-2
The derivitive of 1/2 is 0.
2007-05-11 08:05:15 · answer #9 · answered by Dr D 7 · 1⤊ 1⤋
Most people above gave you right answers. Don't look at the bad ratings.
The reason you did wrong is that
(x/2)' = 1/2 not 1
2007-05-11 08:56:14 · answer #10 · answered by sahsjing 7 · 1⤊ 0⤋