How can I use the power and quotient rule to solve this problem?

[xvtsx]

Hi everyone,
I have been trying to do this problem in both ways but I can't get the same answer the book says. This is the problem:

x/ sqrt (x^2 +1)

With quotient rule I got until the point I have [(x^2 +1)^1/2 - x^2/(x^2 +1)^1/2]/(x^2 +1)
And with power rule I have [1/sqrt(x^2 +1)] - [x^2/(x^2 +1)^3/2]

If you guys can walk me through the problem, it would be nice. Thanks

 

[arildno]

Well, let's look at your result from the quotient rule:
\frac{\sqrt{x^{2}+1}-\frac{x^{2}}{\sqrt{x^{2}+1}}}{x^{2}+1}=\frac{\sqrt{x^{2}+1}}{x^{2}+1}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}=\frac{1}{\sqrt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)^{\frac{3}{2}}}

Does that look familiar?

Furthermore, we may expand our first fraction with the factor (x^{2}+1).
Then, we get:

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

 

[xvtsx]

Can you explain me in details how you combined that part of the fraction because I get lost here \frac{\sqrt {x^{2}+1}}{x^{2}+1}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}=\frac{1}{\sq rt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)^{\frac{3}{2}}}.

 

[cgone]

\frac{\sqrt {x^{2}+1}}{x^{2}+1}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}=\frac{1}{\sq rt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)^{\frac{3}{2}}}

You first multiply so all terms have a common divisor

\frac{\sqrt {x^{2}+1}*\sqrt{x^{2}+1}}{x^{2}+1}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}=\frac{1}{\sq rt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)^{\frac{3}{2}}}

and then simplify

as x^{2}+1 is the same as (x^{2}+1)^{\frac{2}{2} you multiply that with(x^{2}+1)^{\frac{1}{2}

Hope this helps