How can I simplify this calculus problem involving algebraic manipulation?

[Haldhad]

Homework Statement

This is really bugging me it's actually a calculus question, I got to the last step, but couldn't work out how they went from that to the answer.

Homework Equations

The Quotient rule but don't trouble yourself, I am just worried about the algebraic manipulation involved.

The Attempt at a Solution

Well I went from:

(x-2)/(x^3-6x+23)^(3/2)

Using both the chain rule and quotient rule, but this irrelevant. I have the answer, just not sure why it works.

I got:

[1-(3x-2)(2x-6)]/[2(x^2-6x+23)^(5/2)]

And can't seem to work to:

(-2x^2+9x-5)/(x^2-6x+23)^(5/2)

Which comes from that rule as well.

Please move this if its calculus and not just algebra. But this is doing my head in, am I factorising badly. I should know this by now!

It's something I will be marked on and I really don't need to explain the simplification, they give the answer as part of the question btw, so it's understood as true, but for peace of mind!

 

[Ray Vickson]

Homework Statement

This is really bugging me it's actually a calculus question, I got to the last step, but couldn't work out how they went from that to the answer.

Homework Equations

The Quotient rule but don't trouble yourself, I am just worried about the algebraic manipulation involved.

The Attempt at a Solution

Well I went from:

(x-2)/(x^3-6x+23)^(3/2)

Using both the chain rule and quotient rule, but this irrelevant. I have the answer, just not sure why it works.

I got:

[1-(3x-2)(2x-6)]/[2(x^2-6x+23)^(5/2)]

And can't seem to work to:

(-2x^2+9x-5)/(x^2-6x+23)^(5/2)

Which comes from that rule as well.

Please move this if its calculus and not just algebra. But this is doing my head in, am I factorising badly. I should know this by now!

It's something I will be marked on and I really don't need to explain the simplification, they give the answer as part of the question btw, so it's understood as true, but for peace of mind!

If you want the derivative of f(x) = (x-2)/(x^3-6x+23)^(3/2) (as you wrote) then both the results you gave are incorrect. You switched from having "x^3" in the denominator of f(x) to "x^2" in the denominators of your final formulas. What was the correct statement of the original problem?

RGV

 

[Haldhad]

If you want the derivative of f(x) = (x-2)/(x^3-6x+23)^(3/2) (as you wrote) then both the results you gave are incorrect. You switched from having "x^3" in the denominator of f(x) to "x^2" in the denominators of your final formulas. What was the correct statement of the original problem?

RGV

It's supposed to be x^2 in the original (typo my bad) and the actual question is not that important nor is the derivation before simplification as I only want to know how they simplified from the last step to the end step as given. Which is why I put this question in pre calculus. I also forgot to get rid of the 2 lots of the denominator which I canceled in the previous step. I just can't see how or why they simplified it like that. Thus I am feeling rather stupid atm.$$\1-\frac{(3x-2)(2x-6)}{(x^2-6x+23)^{\frac{5}{2}}$$

$$\rightarrow \frac{-2x^2+9x-5}{(x^2-6x+23)^{\frac{5}{2}}$$

I'm not sure why its putting this in that format but meh.

[1-(3x-2)(2x-6)]/(x^2-6x+23)^(5/2) ---->

(-2x^2+9x-5)/(x^2-6x+23)^(5/2)

 

[eumyang]

It's supposed to be x^2 in the original (typo my bad)

Then please fix your original post.

Using both the chain rule and quotient rule, but this irrelevant.

It is relevant, I think, because I'm not getting this expression:
$$\frac{1-(3x-2)(2x-6)}{(x^2-6x+23)^{5/2}}$$

as part of my work in the problem (is this what you're trying to write?). You'll need to show the work involved from the beginning to the expression above.

 

[Sybren]

One of these two expressions is not right, because they do not equal each other:
$$\frac{1-(3x-2)(2x-6)}{2(x^2-6x+23)^{5/2}}\neq\frac{-2x^2+9x-5}{(x^2-6x+23)^{5/2}}.$$

 

[Haldhad]

The time for editing had past or I would of edited.

It won't let me type in latex without posting it in some strange code format for some bizarre reason. And showing the working without it will be needlessly confusing, long winded and ultimately pointless since its clear I've gone wrong somewhere anyway.

The right hand side after the not equal to, is in fact the answer given by the question. I guess I'll just have to run through it again, clearly I have messed up somewhere. Thanks anyway.

EDIT:

$$x^2$$

Very odd it's working now.

I'll have another go and post if I have any problems.

 

[Ray Vickson]

The time for editing had past or I would of edited.

It won't let me type in latex without posting it in some strange code format for some bizarre reason. And showing the working without it will be needlessly confusing, long winded and ultimately pointless since its clear I've gone wrong somewhere anyway.

The right hand side after the not equal to, is in fact the answer given by the question. I guess I'll just have to run through it again, clearly I have messed up somewhere. Thanks anyway.

EDIT:

$$x^2$$

Very odd it's working now.

I'll have another go and post if I have any problems.

You can always post in ASCII. For example, f(x) = (x-2)/(x^2-6*x+23)^(3/2) is clear enough.

RGV

 

[Haldhad]

You can always post in ASCII. For example, f(x) = (x-2)/(x^2-6*x+23)^(3/2) is clear enough.

RGV

It's not so much putting it in anything, it just isn't the shortest question in the world. It would be a little annoying having to read 10 steps to get to the end in ASCII, I know I wouldn't want to. Anyhoo thanks; can do it in latex now, I'll give it another try. Clearly I messed up on something obvious on the way there.