I finding the derivative of this quotient

[realism877]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I got to this (x^2+1)^-1/2[(x^2+1)-x^2/(x^2+1)

 

[lanedance]

i find it easiest to write as a product and just use product rule, which is a good check
<br /> \frac{x}{(x^2+1)^{1/2}}<br /> = x(x^2+1)^{-1/2}<br />

 

[culturedmath]

I don't really understand what you have written as an attempt? (x^2+1) in the beginning is to what power? Have in mind there are some cancellations in the end. You can really use the product rule, most of the times it makes things easier but I believe it is up to you. The other way is to use the Quotient rule or the definition of derivatives using limits. Which part the quotient do you find hard?

Good Luck.

 

[HallsofIvy]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I got to this (x^2+1)^-1/2[(x^2+1)-x^2/(x^2+1)

By the quotient rule, this is
\frac{(x)&#039;(x^2+ 1)^{1/2}- x((x^2+ 1)^{1/2}}{x^2+ 1}
= \frac{(x^2+ 1)^{1/2}- x((1/2)(x^2+ 1)^{-1/2}(2x))}{x^2+ 1}
= \frac{x^+ 1)^{1/2}}{x^2+1}- \frac{x^2}{(x^2+1)^{3/2}}
= \frac{x^2+ 1}{(x^2+ 1)^{3/2}- \frac{x^2}{(x^2+1)^{3/2}}
= \frac{1}{(x^2+1)^{3/2}}

That is what you got except that you forgot the obvious (x^2+ 1)- x^2= 1!

 

[realism877]

how did you get 3/2?

 

[HallsofIvy]

I had x^2+ 1 in the denominator and (x^2+ 1)^{-1/2} in the numerator. Moving that into the denominator,
\frac{(x^2+ 1)^{-1/2}}{x^2+ 1}= \frac{1}{(x^2+1)(x^2+ 1)^{1/2}}= \frac{1}{(x^2+ 1)^{3/2}}

 

[realism877]

I know that I'm making this seem harder than it is, but I'm still confused.

You moved the LaTeX Code: (x^2+ 1)^{-1/2} to the bottom, but I still don't understand how u got 3/2.

 

[dextercioby]

Well, what do you know about muliplying powers ?

 

[realism877]

I know about multiplying powers, but where did you have to multiply powers?

 

[realism877]

I figured it out. Thank you very much.