Find fourth derivative of (-8x)/(1-x)?

Question

I have tried working through all the derivatives but, I think I am getting lost between the 2nd and 3rd. Please help if possible. Thanks

Answer 1

y'=[(1-x)(-8)-(-8x)(-1) ]/(1-x)^2
=[-8+8x-8x ]/(1-x)^2
=[-8 ]/(1-x)^2

y''=-8(-2)(1-x)^{-3}(-1)
=-16(1-x)^{-3}

y'''=(-16)(-3)(1-x)^{-4}(-1)
=-(16)3(1-x)^{-4}

y''''=-(16)3(-4)(1-x)^{-5}(-1)
=-(16)(12)(1-x)^{-5}

Answer 2

Let f(x) = -(8x) / (1 - x). Then,
f'(x) = [(1 - x)(-8) - (-1)(-8x)] / (1 - x)^2
= (-8 + 8x -8x) / (1 - x)^2
= (-8) / (1 - x)^2
= (-8) (1 - x)^(-2).

Thus, the second, third and fourth derivative of f are as follows:
f''(x) = (-8)(-2) (1 - x)^(-3)
f'''(x) = (-8)(-2)(-3) (1 - x)^(-4)
f''''(x) = (-8)(-2)(-3)(-4) (1 - x)^(-5) = 192 / (1 - x)^5 as desired.

Answer 3

f(x) = -8x/(1 - x)

Or,
f(x) = 8x/(x - 1)
-----------------------
The first derivative:
f'(x) = [(x - 1) d/dx (8x) - 8x d/dx (x - 1)]/(x - 1)²

And
f'(x) = [(x - 1)(8) - 8x(1)]/(x - 1)²

f'(x) = (8x - 8 - 8x)/(x - 1)²

f'(x) = -8/(x - 1)²
----------------------------
The second derivative...
f"(x) = -8 d/dx 1/(x - 1)²

And
f"(x) = -8 (- 2(x - 1)/[(x - 1)²]²)

f"(x) = 16/(x - 1)³
-------------------------
The third derivative,
f"'(x) = 16 d/dx 1/(x - 1)³

And
f"'(x) = 16 (-3)(x - 1)²/[(x - 1)³]²

f"'(x) = -48/(x - 1)^4
---------------------------
Lastly, the fourth derivative...
f""(x) = -48 d/dx 1/(x - 1)^4

And
f""(x) = -48 (-4(x - 1)³/[(x - 1)^4]²)

f""(x) = 192/(x - 1)^5

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Answer 4

I would tell you, but I can't understand the equation. Is -3x^4 the numerator and 1-x the denominator?

Answer 5

y = -8x(1-x)^-1
y' = -8(1-x)^-1 -8x(1-x)^-2
y'' = -16(1-x)^-2 -16x(1-x)^-3
y''' = -48(1-x)^-3 -48x(1-x)^-4
y'''' = -192(1-x)^-4 - 192x(1-x)^-5

Answer 6

You should go to the 4th. The answer is popcorn.