f''(x) Trouble - 2nd Derivative of f(x) = (x^2) / (x^2 + 3)

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2000WS6

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Part of a more complex problem, but my issue is with this final portion which is finding the 2nd derivative of this equation. The solutions manual seems to simplify the equaiton in a manner in which I can't understand for the life of me. Without those last simplifications, it's almost impossible for me to take the 2nd derivative.

problem: f(x) = (x^2) / (x^2 + 3)

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

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

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

f'(x) is never undefined....

Critical number:
6x = 0 -> 0 is my critical number

Test Points:
f'(-1) = -.375
f'(1) = .375

Abs. min. @ (0,0)

2nd Derivative....

f''(x) = (((x^2 + 3)^2) (6)) - ((6x) (2 (x^2 + 3) (2x)))) / (x^2 + 3)^4

... it all just falls apart from there - the solution manual goes from there as:
f''(x) = (((x^2 + 3) (6)) - ((6x) (2x))) / (x^2 + 3)^3

f''(x) = (18(1 - x^2)) / (x^2 + 3)^3


The first part isn't that big of a jump for me but if someone could help me understand what exactly is 'going away' and WHY, I'd appreciate it. By the time I look at the final answer the book has, I'm in la-la land lol. Thanks for the help!
 
Re: f''(x) Trouble - 2nd Derivative

2000WS6 said:
Part of a more complex problem, but my issue is with this final portion which is finding the 2nd derivative of this equation. The solutions manual seems to simplify the equaiton in a manner in which I can't understand for the life of me. Without those last simplifications, it's almost impossible for me to take the 2nd derivative.

problem: f(x) = (x^2) / (x^2 + 3)

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

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

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

f'(x) is never undefined....

Critical number:
6x = 0 -> 0 is my critical number

Test Points:
f'(-1) = -.375
f'(1) = .375

Abs. min. @ (0,0)

2nd Derivative....

f''(x) = (((x^2 + 3)^2) (6)) - ((6x) (2 (x^2 + 3) (2x)))) / (x^2 + 3)^4

f''(x) = [6(x[sup:1nn7qbbp]2[/sup:1nn7qbbp] + 3)[sup:1nn7qbbp]2[/sup:1nn7qbbp] - 24x[sup:1nn7qbbp]2[/sup:1nn7qbbp](x[sup:1nn7qbbp]2[/sup:1nn7qbbp] + 3)]/(x[sup:1nn7qbbp]2[/sup:1nn7qbbp] + 3)[sup:1nn7qbbp]4[/sup:1nn7qbbp]

f''(x) = 6(x[sup:1nn7qbbp]2[/sup:1nn7qbbp] + 3)[(x[sup:1nn7qbbp]2[/sup:1nn7qbbp] + 3) - 4x[sup:1nn7qbbp]2[/sup:1nn7qbbp]]/(x[sup:1nn7qbbp]2[/sup:1nn7qbbp] + 3)[sup:1nn7qbbp]4[/sup:1nn7qbbp]

f''(x) = 6(3 - 3x[sup:1nn7qbbp]2[/sup:1nn7qbbp])/(x[sup:1nn7qbbp]2[/sup:1nn7qbbp] + 3)[sup:1nn7qbbp]3[/sup:1nn7qbbp]

f''(x) = 18(1 - x[sup:1nn7qbbp]2[/sup:1nn7qbbp])/(x[sup:1nn7qbbp]2[/sup:1nn7qbbp] + 3)[sup:1nn7qbbp]3[/sup:1nn7qbbp]


... it all just falls apart from there - the solution manual goes from there as:
f''(x) = (((x^2 + 3) (6)) - ((6x) (2x))) / (x^2 + 3)^3

f''(x) = (18(1 - x^2)) / (x^2 + 3)^3


The first part isn't that big of a jump for me but if someone could help me understand what exactly is 'going away' and WHY, I'd appreciate it. By the time I look at the final answer the book has, I'm in la-la land lol. Thanks for the help!
Click to expand...
 
Re: f''(x) Trouble - 2nd Derivative

The algebra is the most troublesome part of calc. Sometimes the manuals simplify to the point you don't understand.

I will start with taking the derivative of the first derivative.

Let's rewrite the first derivative as \(\displaystyle 6x(x^{2}+3)^{-2}\), then use the product rule:

\(\displaystyle 6x(x^{2}+3)^{-2}\)

\(\displaystyle 6x(-2)(x^{2}+3)^{-3}(2x)+(6)(x^{2}+3)^{-2}\)

Follow so far?.

\(\displaystyle -24x^{2}(x^{2}+3)^{-3}+6(x^{2}+3)^{-2}\)

The exponents in the \(\displaystyle (x^{2}+3)\) are negative so put them in the denominator to make them positive.
\(\displaystyle \frac{-24x^{2}}{(x^{2}+3)^{3}}+\frac{6}{(x^{2}+3)^{2}}\)

Now, in order to add, the denominators have to be the same. Multiply top and bottom of the right one by \(\displaystyle (x^{2}+3)\)

\(\displaystyle \frac{-24x^{2}}{(x^{2}+3)^{3}}+\frac{6}{(x^{2}+3)^{2}}\cdot\frac{(x^{2}+3)}{(x^{2}+3)}\)

\(\displaystyle \frac{-24x^{2}}{(x^{2}+3)^{3}}+\frac{-18x^{2}+18}{(x^{2}+3)^{3}}\)

Add across the top and factor out the 18:

\(\displaystyle \frac{-18(x^{2}-1)}{(x^{2}+3)^{3}}\)

\(\displaystyle \frac{18(1-x^{2})}{(x^{2}+3)^{3}}\)

See?.
 
Yes, I do see. Thank you very much. I guess I just have to look harder for simplifications. My algebra isn't the best, that I know.

One more question - I've looked at the formatting write-ups but how did you get that format on your text, galactus? I'm usually posting from a Mac, FYI.
 
That's called LaTex. Click on 'quote' at the upper right corner of my post to see the code I typed in to display like that.
 
galactus said:
That's called LaTex. Click on 'quote' at the upper right corner of my post to see the code I typed in to display like that.
Click to expand...

Interesting. Time for me to start learning. Thanks.
 
Give something a practice run and see of it works.
 
galactus said:
Give something a practice run and see of it works.
Click to expand...

Hmm... practice run I'll give both a practice run - the script and the algebra

Having another problem - don't know which way to go. In its current form, I can easily get the critical points, but I will have to take the 2nd derivative as well, which will be very hard as the problem stands. Which way do I go (distribute the 2x, or leave it in its current state) and WHY - I need to be able to make this decision on my own. Thanks

This is currently the 1st derivative (mid-way through simplification):

\(\displaystyle \frac{2x[(x^{2}-9)-(x^{2}+1)]}{(x^{2}-9)^{2}}\)

... not too hard at all... especially for someone versed in C and C++ ;)
 
Where did the x^2-9 come from?.

\(\displaystyle f(x)=\frac{x^{2}}{x^{2}+3}\)

Quotient rule:

\(\displaystyle f'(x)=\frac{(x^{2}+3)(2x)-(x^{2})(2x)}{(x^{2}+3)^{2}}\)

\(\displaystyle f'(x)=\frac{2x^{3}+6x-2x^{2}}{(x^{2}+3)^{2}}\)

\(\displaystyle f'(x)=\frac{6x}{(x^{2}+3)^{2}}\)

2nd derivative using quotient rule(you could also use the product rule).

\(\displaystyle f''(x)=\frac{(x^{2}+3)^{2}(6)-6x(2)(x^{2}+3)(2x)}{(x^{2}+3)^{4}}\)

I will let you simplify. Okey-doke.
 
It was there in the 1st deriv. (too tired to look for it right now... or think for that matter. It's 5:20 am!) - I ended up figuring it out after about 20 minutes of thinking through it. Ended up being pretty friendly once I put some deep thought into it. You've been more than helpful, galactus. Thank you.
 
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