Calculus Question

[Weyoun]

Hi all

Yesterday I was given 2 pages of calculus to complete for our maths class, and came upon a question that needs a very sloppy solution. The question asks me to find the derivative of f(x) = x(x^2 + 1)^(-1/2) I have tried the quotient and product rule for this question, and end up with the correct answer, only after some dodgey substitution. I have no desire to have to rely upon such methods come exam time. The textbook states the derivative as (x^2 + 1)^(-3/2) I know the parentheses are horrible to read, but here goes...

Here's my working:
where f(x) = U.V
f(x) = x/(x^2 + 1)^(1/2)

U = x du/dx = 1

V = (x^2 + 1)^(-1/2) dv/dx = -1/2 * (x^2 + 1)^(-3/2) * 2x
. = -x(x^2 + 1)^(-3/2)

Using the product rule
f'(x) = V*du/dx + U*dv/dx
. = (x^2 + 1)^(-1/2) -x^2(x^2 + 1)^(-3/2)
. = ((x^2 + 1)^(1/2) * (x^2 + 1)^(-1/2) - x^2(x^2 + 1)^(-3/2) * (x^2 + 1)^(1/2))/(x^2 + 1)^(1/2)
. = (1 - x^2(x^2 + 1)^-1)/(x^2 + 1)^(1/2)
. = 1/(x^2 + 1)^(1/2) - x^2/(x^2 + 1)^(3/2)
. = (x^2 + 1)/(x^2 + 1)^(3/2) - x^2/(x^2 + 1)^(3/2)
. = (x^2 + 1 - x^2)/(x^2 + 1)^(3/2)
. = 1/(x^2 + 1)^(3/2)
. = (x^2 + 1)^(-3/2)

What I don't like doing is having to multiply throughout by (x^2 + 1)^(1/2) / (x^2 + 1)^(1/2) to remove the first term, as it will most likely not work for other questions (of course changing the multiplication factors where necessary). This problem most likely arises from my lack of knowledge with regard to indicies, and am probably just missing a single factorisation... I'm only new to calculus (2 weeks), so by any means feel free to ridicule/make fun of/correct me where necessary

*Thanking you in advance*

 

[Weyoun]

*bump*

 

[yiwonder]

It's actually quite simple. I don't know exactly how your teacher taught you to do this. I also assume that you know the power rule and chain rule along with the product rule.

f(x)=x(x^2+1)^(-1/2)
Here is the use of the product rule:
1(x^2+1)^(-1/2)+(-1/2)x(x^2+1)^(-3/2)

BUT, using the power rule on anything but a monomial is illegal. Therefore, we must use the chain rule. The chain rule tells us to take the derivative of the binomial that we just used the power rule on. This means find the derivative of (x^2+1) and multiply it to that part of the product rule. See below.
1(x^2+1)^(-1/2)+(-1/2)x(x^2+1)^(-3/2)(2x)

Now, we just need to clean it up.
(x^2+1)^(-1/2)-x^2(x^2+1)^(-3/2)

If you want, you can clean it up some more to get the same answer you got, but we can just leave it in this form.

Do you understand or are you just completely confused? If you're confused, I'll take it slower. The easiest (the only way that I know of) way to do this is to use the power rule, the product rule, and the chain rule.

 

[Weyoun]

thanks for the reply yiwonder

We have been taught those rules, although we didn't use that exact terminology. We were taught the derivative of x^n for n rational, and its application for functions of functions. If you look at line 4 of my proof, you'll find I have exactly what you wrote, with the exception of taking the negative indices into account as denominators My initial problem was converting that back into its simplest form - what was written in the textbook. I recieved a PM from Haircut with regard to this question, who gave me an insight into factorisation. Instead of taking out a factor of (x^2 + 1)^(-1/2), I took out (x^2 + 1)^(-3/2), removing several lines from my proof. My amended working:

Using the product rule and x^n rule where appropriate
f(x) = x(x^2 + 1)^(-1/2)
U = x
du/dx = 1
V = (x^2 + 1)^(-1/2)
dv/dx = -1/2 * (x^2 + 1)^(-3/2) * 2x
. . . . . = -x(x^2 + 1)^(-3/2)

dy/dx = (x^2 + 1)^(-1/2) - x^2(x^2 + 1)^(-3/2)
. . . . . = (x^2 + 1)^(-3/2) * (x^2 + 1 - x^2) . . . . . . . . . . . . . NB: multiplication of indices is addition (-3/2 + x = -1/2, x = 1)
. . . . . = 1/(x^2 + 1)^(3/2)

Thankyou for your reply Any more comments would be much appreciated

 

[Renob]

1+1=2
2+2=4
3+3=6
4+4=8

I hope that helps

 

[yiwonder]

Yeah, I saw that you had the beginnings, but it seemed like you went through a lot of extra work. It took me 3 lines of work. Add another couple lines if you want to get it to the answer you gave. I see what you did now. It sure is a lot of work to simplify that isn't it?

We don't have to simplify those beyond where I went because usually, you're going to but numbers in for x. It gets to a point where the algebra to simplify would take much longer than just putting in the number and going from there.

 

[AMDPwred]

I hope I never have to take Calc. Oh wait, yeah I will, I want to be a Computer Science major

 

[pillage2001]

<< I hope I never have to take Calc. Oh wait, yeah I will, I want to be a Computer Science major >>



It's really......not that hard.

 

[AMDPwred]

lol, right...

 

[Weyoun]

Seriously, the hardest part of Calculus is knowing your indices and work on functions. The actual Calculus is relatively easy You'd also be surprised at what you can do with it. We were given this problem not so long ago. You have a 10 * 5 sheet of cardboard. By cutting out 4 squares of equal area, it is possible to create a topless box. Find the size of the square that gives the largest area for the box. Doing these seemingly impossible questions is now relatively easy with calculus