#### allegansveritatem

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Good effort. Don't be afraid to write things over, rather than messily crossing things out.

When you multiplied by the (6x+1)^(2/3), you did that only in the numerator. Whoops. Multiply by that factor in the denominator, too. You should have been suspicious about this when the fractional exponents just vanished. That IS too good to be true.

Note: When you are in calculus, this expression will make a lot more sense. It's not just a bunch of symbols thrown into a fan.

When you multiplied by the (6x+1)^(2/3), you did that only in the numerator. Whoops. Multiply by that factor in the denominator, too. You should have been suspicious about this when the fractional exponents just vanished. That IS too good to be true.

Note: When you are in calculus, this expression will make a lot more sense. It's not just a bunch of symbols thrown into a fan.

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Where did the (2) come from on the second line?

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I made a mistake in copying the original problem. If you look at my original post you will see it. In the problem. as it should have been transcribed, there was a 1/3 and a 6 as factors. In my latest post I overlooked the 6 but the 2 in the rest of the work "includes" the 6, so the work is correct. At the time I posted I did not know it was correct because I was using my solutions manual to check things and the solutions manual only gives solutions for odd numbered problems. But today, after again working this out and getting same answer, I suddenly recalled that my text gives answers for all the problems for review sections (which is where this problem comes from) so I checked it in the book and the book confirmed my result. And what is strange is when I try both the original expression and the simplified expression in my calculator (after I have assigned x a value) I get two different answers. the difference is fairly slight but significant, I would say. How to explain this?Where did the (2) come from on the second line?

Yes, it does look nasty, but tkhunny is correct that this sort of ugliness arises in calculus. Technically, the expression above is the derivative ofI am asked to simplify this:

View attachment 12199

Here is what I got...and this is only the latest in a series:

'View attachment 12200

I am full of doubt about this because...why would they have that (4-x^2) in both numerator and denominator if it wasn't meant to disappear somehow? Anyway, this is my best shot so far.

\(\displaystyle \dfrac{(6x + 1)^{1/3}}{4 - x^2}.\)

The derivative of a function is the function's "slope."

What I would do

\(\displaystyle \dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6) - (6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = \)

\(\displaystyle \dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6)}{(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = \)

\(\displaystyle \dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = \)

\(\displaystyle \dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{2/3} * (6x + 1)^{1/3}(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} = \)

\(\displaystyle \dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} = \)

\(\displaystyle \dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{-\ 12x^2 - 2x}{(6x + 1)^{2/3}(4 - x^2)^2} = \)

\(\displaystyle \dfrac{8 - 2x^2 + 12x^2 + 2x}{(6x + 1)^{2/3}(4 - x^2)^2} = \dfrac{10x^2 + 2x + 8}{(6x + 1)^{2/3}(4 - x^2)^2}\)

I shall test it using x = 0.

\(\displaystyle \dfrac{(4 - 0^2)\left( \frac{1}{3} \right )(6 * 0 + 1)^{-2/3}(6) - (6 * 0 + 1)^{1/3}(-\ 2 * 0)}{1^{2/3} * (4 - 0^2)^2} = \dfrac{2(4)(1)^{-2/3}}{1 * 4^2} = \dfrac{1}{2}.\)

\(\displaystyle \dfrac{10 * 0^2 + 2 * 0 + 8}{(6 * 0 + 1)^{2/3}(4 - 0^2)^2} = \dfrac{8}{1 * 16} = \dfrac{1}{2}.\)

Your algebra looks good to me. What was your test number on the calculator?

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well, I just tried it again and this time the expressions produced the same result...so I must have made some kind of mistake when I input the data to the calculator. So...a happy ending. Thanks for checking it.Yes, it does look nasty, but tkhunny is correct that this sort of ugliness arises in calculus. Technically, the expression above is the derivative of

\(\displaystyle \dfrac{(6x + 1)^{1/3}}{4 - x^2}.\)

The derivative of a function is the function's "slope."

What I would doinitiallyis to break the fraction into two expressions that will be a bit easier to work with, then simplify, then get common denominators, simplify again, and add. Trying to work on that mess in one fell swoop will drive you bonkersquickly.

\(\displaystyle \dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6) - (6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = \)

\(\displaystyle \dfrac{(4 - x^2)\left( \frac{1}{3} \right )(6x + 1)^{-2/3}(6)}{(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = \)

\(\displaystyle \dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{1/3}(-\ 2x)}{(4 - x^2)^2} = \)

\(\displaystyle \dfrac{2(4 - x^2)}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)^{2/3} * (6x + 1)^{1/3}(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} = \)

\(\displaystyle \dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{(6x + 1)(-\ 2x)}{(6x + 1)^{2/3}(4 - x^2)^2} = \)

\(\displaystyle \dfrac{8 - 2x^2}{(6x + 1)^{2/3}(4 - x^2)^2} - \dfrac{-\ 12x^2 - 2x}{(6x + 1)^{2/3}(4 - x^2)^2} = \)

\(\displaystyle \dfrac{8 - 2x^2 + 12x^2 + 2x}{(6x + 1)^{2/3}(4 - x^2)^2} = \dfrac{10x^2 + 2x + 8}{(6x + 1)^{2/3}(4 - x^2)^2}\)

I shall test it using x = 0.

\(\displaystyle \dfrac{(4 - 0^2)\left( \frac{1}{3} \right )(6 * 0 + 1)^{-2/3}(6) - (6 * 0 + 1)^{1/3}(-\ 2 * 0)}{1^{2/3} * (4 - 0^2)^2} = \dfrac{2(4)(1)^{-2/3}}{1 * 4^2} = \dfrac{1}{2}.\)

\(\displaystyle \dfrac{10 * 0^2 + 2 * 0 + 8}{(6 * 0 + 1)^{2/3}(4 - 0^2)^2} = \dfrac{8}{1 * 16} = \dfrac{1}{2}.\)

Your algebra looks good to me. What was your test number on the calculator?