Help combining Fractions?

Question

the original problem deals w/ finding the derivative of
F(x)= 4X^(1/6) -3x^(-1/6) +3

ok, no problems so far...i go in, apply the power rule, and get

4/6 X^(-5/6) + 3/6 X^(-7/6)

the thing is, im having trouble combining them, in my answers i get numerators in the top, like 3+4 (all numbers), however all the "correct" answer choices have "X" in the top AND in the bottom.

any help?

-confused-

Answer 1

The original problem deals w/ finding the derivative of
F(x)= 4X^(1/6) -3x^(-1/6) +3
F'(x)= 4/6 X^(-5/6) + 3/6 X^(-7/6)

It would help if you posted the choices.

If they say the answer *must have* integral powers of X only, this can't be right, in general you can't reduce a polynomial with non-integral powers of X to a quotient of polynomials with only integral powers of X.
Unless they asked for a Taylor series approximation to the result, truncated to a finite number of terms?
For example you can rewrite things as follows:
F'(x)= 4/6 X^(-5/6) + 3/6 X^(-7/6)
= (4. X^(1/6) + 3. X^(-1/6)) / (6X) [integral-power denominator]
or as
= (3 + 4. X^(1/3)) / (6X^(7/6))
= (1 + (4/3). X^(1/3)) / (2X^(7/6))

Answer 2

Your answer is correct.

You could rewrite it as 2/(3x^5/6)+ 1/(3X^7/6)

Some people don't like negative exponents in the final answer.
I can't see any way to have x in both the numerator and denominator.

Answer 3

You have (4/(6x^(5/6))) + (3/(6x^(7/6))) so mutiply top and bottom of the first term by x^(7/6) and multiply top and bottom of the 2'nd term by x^(5/6) to get a common denominator of 6x² and add the numerator terms to get
((4x^(-7/6)) + (3x^(-5/6)))/x²


Doug

Answer 4

4/6 X^(-5/6) + 3/6 X^(-7/6)

= 2/3 X^(-5/6) + 1/2 X^(-7/6)

Take {X^(-1/6)}/6 common, then

[{X^(-1/6)}/6]*[4*X + (3/X)]

Look, X^(-7/6) = X^( -1 - 1/6) = X^(-1) * X^(-1/6) = (1/X) * X^(-1/6)

Again, X^(-5/6) = X^(1 - 1/6) = X^1 * X^(-1/6)