• May 23rd 2010, 01:19 AM
dwatkins741

((x^2+1)^(1/2) - x^2(x^2-1)^(-1/2)) / (x^2-1)

The book says 1/(x^2+1)^3/2

While trying different things I figured out the rule that the square root of a divided by a = 1/ square root of a. There must a rule I could use here. My answer always has x^2 in the numerator.

Thanks!
• May 23rd 2010, 02:06 AM
Sudharaka
Quote:

Originally Posted by dwatkins741

((x^2+1)^(1/2) - x^2(x^2-1)^(-1/2)) / (x^2-1)

The book says 1/(x^2+1)^3/2

While trying different things I figured out the rule that the square root of a divided by a = 1/ square root of a. There must a rule I could use here. My answer always has x^2 in the numerator.

Thanks!

Dear dwatkins741,

So the book says that,

$\frac{(x^2+1)^{\frac{1}{2}}-x^2(x^2-1)^{-\frac{1}{2}}}{x^2-1}=\frac{1}{(x^2+1)^{3/2}}$

But notice that if x=0 then, $\frac{(x^2+1)^{\frac{1}{2}}-x^2(x^2-1)^{-\frac{1}{2}}}{x^2-1}=-1$

But $\frac{1}{(x^2+1)^{3/2}}=1$

Hence there is a mistake in this problem,

$\frac{(x^2+1)^{\frac{1}{2}}-x^2(x^2-1)^{-\frac{1}{2}}}{x^2-1}\neq\frac{1}{(x^2+1)^{3/2}}$

• May 23rd 2010, 02:12 AM
dwatkins741
Correction
I sincerely apologize. I had two signs wrong when I typed the problem. Please take another look at it.

((x^2+1)^(1/2) - x^2(x^2+1)^(-1/2)) / (x^2+1)

Sorry, thanks.
• May 23rd 2010, 02:26 AM
Sudharaka
Quote:

Originally Posted by dwatkins741
I sincerely apologize. I had two signs wrong when I typed the problem. Please take another look at it.

((x^2+1)^(1/2) - x^2(x^2+1)^(-1/2)) / (x^2+1)

Sorry, thanks.

Dear dwatkins741,

Don't mention it. Now you can solve the problem. First multiply the denominatior and the numerator by, $(x^2+1)^{\frac{1}{2}}$. Hope you can continue.
• May 23rd 2010, 05:06 AM
Wilmer
Quote:

Originally Posted by dwatkins741
((x^2+1)^(1/2) - x^2(x^2+1)^(-1/2)) / (x^2+1)

Go this way: let a = x^2 + 1 ; then:
[sqrt(a) - x^2/sqrt(a)] / a : remember that k^(-p) = 1/k^p
= [sqrt(a)sqrt(a) - x^2] / [a sqrt(a)]
= (a - x^2) / [a^1 a^(1/2)]
= (a - x^2) / a^(3/2) : remember that k^p k^q = k^(p+q)

Finish it by substituting back in...