# Basic calculus Quotient rule question

• Nov 21st 2008, 07:50 PM
silverbird
Basic calculus Quotient rule question
Hi iam doin quotient rules in calculus and iam just getting stuck at a certain stage in this question.
The question is
$\displaystyle = \frac{3x+1}{\sqrt{x+1}}$
and the answer they have down is
$\displaystyle =\frac{3x+5}{2\sqrt{(x+1)^3}}$
and iam using the forumula
$\displaystyle y=\frac{vu'-uv'}{v^2}$
and iam getting stuck at
$\displaystyle \frac{3(x+1)^\frac{1}{2}-\frac{1}{2}(x+1)^{-\frac{1}{2}}(3x+1)}{x+1}$
could someone please show me the working out and how you do this question thx
• Nov 21st 2008, 08:03 PM
whipflip15
Multiply tnumerator and denominator by $\displaystyle 2(x+1)^{1/2}$ and note that $\displaystyle \sqrt{(x+1)^3}=(x+1)^{3/2}.$
• Nov 21st 2008, 10:58 PM
silverbird
Still a little confused would anybody be able to show me the working out steps please. Whipflip i can kind of see how that gets the right answer but why do you do that?
• Nov 21st 2008, 11:25 PM
o_O
$\displaystyle \frac{\displaystyle 3(x+1)^{\frac{1}{2}} \ - \ \frac{1}{2} \cdot \frac{1}{(x+1)^{\frac{1}{2}}} \cdot (3x+1)}{(x+1)^1}$

Notice that if we follow whipflip's advice, multiplying through with $\displaystyle \frac{(x+1)^{\frac{1}{2}}}{(x+1)^{\frac{1}{2}}}$ will give us that $\displaystyle (x+1)^{\frac{3}{2}}$ we wanted in the denominator:
$\displaystyle \frac{\displaystyle 3(x+1)^{\frac{1}{2}} \ - \ \frac{1}{2} \cdot \frac{1}{(x+1)^{\frac{1}{2}}} \cdot (3x+1)}{(x+1)^1} \cdot {\color{blue}\frac{(x+1)^{\frac{1}{2}}}{(x+1)^{\fr ac{1}{2}}}}$

$\displaystyle = \frac{\displaystyle 3(x+1)^{\frac{1}{2}}{\color{blue}(x+1)^{\frac{1}{2 }}} - \frac{1}{2} \cdot \frac{{\color{blue}(x+1)^{\frac{1}{2}}}}{(x+1)^{\f rac{1}{2}}} \cdot (3x+1)}{(x+1)^1{\color{blue}(x+1)^{\frac{1}{2}}}}$ .... (Simply multiplied in the blue factor)

$\displaystyle = \frac{\displaystyle 3(x+1)^1 - \frac{1}{2} \cdot {\color{blue}(1)} \cdot (3x+1)}{(x+1)^{\frac{3}{2}}}$ .... (Used the fact that: $\displaystyle a^na^m = a^{n+m}$ and simplified)

$\displaystyle = \cdots$