Kay H. answered • 11/29/13

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Robert's answer looks good. If, however, the 3(2-x) is in the denominator, then try this:

f(x) = x^2 / (6-3x)

Use Chain Rule where a = x^2 and b = (6-3x)^-1

f'(x) = ab' + ba'

f'(x) = (x^2)(-1)(-3)(6-3x)^-2 + 2x(6-3x)^-1

f'(x) = (3x^2)/(6-3x)^2 + (2x)/(6-3x)

Now multiply top and bottom of second term by (6-3x) and simplify:

f'(x) = (3x^2)/(6-3x)^2 + (2x)(6-3x)/(6-3x)^2

Combine terms now that we have a common denominator:

f'(x) = (3x^2 + 12x - 6x^2) / (6-3x)^2

f'(x) = (3x)(4-x)) / (3(2-x))^2)

f'(x) = (3x(4-x)) / (9(2-x)^2)

f'(x) = x(4-x) / (3(2-x)^2)

For the second derivative, now factor out the (1/3) and use chain rule again where a = (4x-x^2) and b = (2-x)^-2

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

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

Multiply terms and simplify:

f''(x) = (1/3) [ (8x-2x^2) / ((2-x)^3) + (4-2x) / ((2-x)^2) ]

To simplify further, multiply top and bottom of the second term by (2-x) and combine terms:

f''(x) = 8 / (3(2-x)^3)

Happy Thanksgiving!