# Find g'(x) By Quotient Rule

• November 12th 2013, 06:37 PM
nycmath
Find g'(x) By Quotient Rule
See picture.
• November 12th 2013, 07:11 PM
SlipEternal
Re: Find g'(x) By Quotient Rule
\begin{align*}g'(x) & = \dfrac{(x+1)\dfrac{-1}{\sqrt{1-x^2}} - \cos^{-1}(x)}{(x+1)^2} \\ & = -\dfrac{1}{(x+1)\sqrt{1-x^2}} - \dfrac{\cos^{-1}(x)}{(x+1)^2}\end{align*}
• November 13th 2013, 05:21 AM
nycmath
Re: Find g'(x) By Quotient Rule
Ok but how did you get to the final answer? Can you break down the complex fraction for me? Can you demonstrate why the book's answer is right?
• November 13th 2013, 05:27 AM
SlipEternal
Re: Find g'(x) By Quotient Rule
$\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c}$

This means

$\dfrac{(x+1)\dfrac{-1}{\sqrt{1-x^2}}-\cos^{-1}(x)}{(x+1)^2} = -\dfrac{(x+1)\dfrac{1}{\sqrt{1-x^2}}}{(x+1)^2} - \dfrac{\cos^{-1}(x)}{(x+1)^2}$

In the first fraction, multiply top and bottom by $\dfrac{\sqrt{1-x^2}}{x+1}$:

$-\dfrac{(x+1)\dfrac{1}{\sqrt{1-x^2}}}{(x+1)^2}\cdot \dfrac{\dfrac{\sqrt{1-x^2}}{x+1}}{\dfrac{\sqrt{1-x^2}}{x+1}} - \dfrac{\cos^{-1}(x)}{(x+1)^2} = -\dfrac{1}{(x+1)\sqrt{1-x^2}} - \dfrac{\cos^{-1}(x)}{(x+1)^2}$
• November 13th 2013, 05:35 AM
nycmath
Re: Find g'(x) By Quotient Rule
I see what to do now. It's not calculus that is giving me a hard time. Calculus is done is the first two or three steps. It is the algebra to break down the problem that is truly the challenge.
• November 13th 2013, 05:57 AM
SlipEternal
Re: Find g'(x) By Quotient Rule
There are lots of strategies for simplifying expressions. At this point, I do it by rote. When I first learned, though, I made a lot of careless mistakes. But, with practice, it became second nature. So, I would recommend finding an intermediate algebra book (any book used for an Algebra II class) and practicing the problems. If you do even one problem a day for a few weeks, I suspect you will see a marked improvement in your algebra skills.
• November 13th 2013, 06:50 AM
nycmath
Re: Find g'(x) By Quotient Rule
I will do as you suggest.