Dervitative Help: f(x) = (x+2)^3 / 4(x-x^2)

[chantelw_12]

I'm stuck on this question...
f(x) = (x+2)^3 / 4(x-x^2)
And I get f'(x) = (x+2)(-x^2 + 6x - 2) / 4(x-x^2)^2.
I'm supposed to get f''(x) = (x+2)(19x^2 - 14x +4) / 2(x-x^2)^3 but I can't seem to arrive at the answer.
Which part of the question am I going wrong in? And how do I go about solving it then?

 

[stapel]

f(x) = (x+2)^3 / 4(x-x^2)
And I get f'(x) = (x+2)(-x^2 + 6x - 2) / 4(x-x^2)^2.
I'm supposed to get f''(x) = (x+2)(19x^2 - 14x +4) / 2(x-x^2)^3 but I can't seem to arrive at the answer.
Which part of the question am I going wrong in? And how do I go about solving it then?

Since you don't provide the instructions, we don't know what is expected. Since you show only the final result of taking the first derivative, we don't know what you did or where you might have gone wrong. But the fact that "the answer" is a second derivative suggests that perhaps you aren't following the instructions, at the very least...?

Please reply with the missing information (the instructions and a clear listing of all of your work and reasoning). Please be complete. Thank you!

Eliz.

 

[chantelw_12]

The instructions ask you to find prove that f(x) = (x+2)^3 / 4(x-x^2) equals f''(x) = (x+2)(19x^2 - 14x +4) / 2(x-x^2)^2.
I have done the problem multiple ways trying to figure out where I keep going wrong.

f(x) = (x+2)^3 / 4(x-x^2)
f'(x) = (1/4)[(3(x+2)^2 * (x-x^2) - (x+2)^3 * (1-2x)) / (x-x^2)^2]
f'(x) = ((x+2)/4)[(3(x-x^2) - (x+2)(1-2x) / (x-x^2)^2]
f'(x) = ((x+2)/4)[(-x^2 +6x -2) / (x-x^2)^2]
f'(x) = (x+2)(-x^2 +6x -2) / 4(x-x^2)^2

f"(x) = (1/4)[(((-x^2 +6 -2) + (x+2)(-2x+6))*(x-x^2) - (x+2)(-x^2 +6x -2)(2)(x-x^2)(1-2x)) / ((x-x^2)^2)^2]
f"(x) = (1/4)(x-x^2)[((-2x^2 +8x +10)(x-x^2) + (2x^2 - 12x +4 -4x^3 + 24x^2 -8x)) / ((x-x^2)^4)]
f"(x) = (1/4)[(-2x^3 +8x^2 +10x +2x^4 -8x^3 - 10x^2 -4x^3 +26x^2 -4x +4) / ((x-x^2)^3)]
f"(x) = (1/4)(2)[(x^4 -7x^3 +12x^2 +3x +2) / ((x-x^2)^3)]
f"(x) = (x^4 -7x^3 +12x^2 +3x +2) / (2(x-x^2)^3)

 

[stapel]

The instructions ask you to find prove that f(x) = (x+2)^3 / 4(x-x^2) equals f''(x) = (x+2)(19x^2 - 14x +4) / 2(x-x^2)^2.

Really? The instructions said to try to show that two functions which are clearly not the same are actually somehow equal to each other...? Well, it can't be done, because they simply are NOT the same.

But you appear, I think, to be instead trying to show that the second derivative of the given function f(x) is equal to the proposed formulation. In case the instructions were actually to "find the second derivative", then:

f(x) = (x+2)^3 / 4(x-x^2)
f'(x) = (1/4)[(3(x+2)^2 * (x-x^2) - (x+2)^3 * (1-2x)) / (x-x^2)^2]
f'(x) = ((x+2)/4)[(3(x-x^2) - (x+2)(1-2x) / (x-x^2)^2]

Did you lose a copy of x + 2 between your second and third line...?

Eliz.

 

[chantelw_12]

No sorry, that was a typo. The instructions are in French and literally translated it means to prove it but we have been taught to find f"(x) to prove it.

f(x) = (x+2)^3 / 4(x-x^2)
f'(x) = (1/4)[(3(x+2)^2 * (x-x^2) - (x+2)^3 * (1-2x)) / (x-x^2)^2]
f'(x) = ((x+2)^2/4)[(3(x-x^2) - (x+2)(1-2x) / (x-x^2)^2]
f'(x) = ((x+2)^2/4)[(-x^2 +6x -2) / (x-x^2)^2]
f'(x) = (x+2)^2(-x^2 +6x -2) / 4(x-x^2)^2