# Simplification of a derivative

• Mar 9th 2010, 06:43 AM
dbakeg00
Simplification of a derivative
I'm in the process of a derivative and I'm having trouble simplifying. Would someone mind helping me out with where to start.

Here is the original problem I was told to differentiate:

$\theta = \frac{1-a\sqrt[3]{t^2}}{1+a\sqrt{t^3}}$

I decided to use the quotient rule and came up with this as the first step:

$\frac{1-at^{\frac{2}{3}}}{ 1+at^{\frac{3}{2}}}$

and then this:

$\frac{(1+at^{\frac{3}{2}})(\frac{-2}{3}at^{\frac{-1}{3}}) - (1-at^{\frac{2}{3}})(\frac{3}{2}at^{\frac{1}{2}})}{(1 +at^{\frac{3}{2}})^2}$

and here is what I multiplied it all out to:

$\frac{\frac{-2}{3}at^\frac{-1}{3} + \frac{-2}{3}a^2t^{\frac{7}{6}} - \frac{3}{2}at^{\frac{1}{2}} - \frac{3}{2}a^2t^{\frac{7}{6}}}{(1+at^{\frac{3}{2}} )^2}$

Now, its been a while and my algebra is rusty so I'm stuck here. What is the first thing you would do to go about simplifying? Thanks in advance.
• Mar 9th 2010, 06:56 AM
Quacky
http://www.mathhelpforum.com/math-he...a638e2e5-1.gif

Yuck.(Surprised) You can certainly take out a factor of a.
• Mar 9th 2010, 08:08 AM
hameerabbasi
Rationalize
Rationalize before differentiating:

$\begin{array}{l}
\theta = \frac{{1 - a\sqrt[3]{{{t^2}}}}}{{1 + a\sqrt {{t^3}} }} \cdot \frac{{1 - a\sqrt {{t^3}} }}{{1 - a\sqrt {{t^3}} }} \\
= \frac{{\left( {1 - a\sqrt[3]{{{t^2}}}} \right)\left( {1 - a\sqrt {{t^3}} } \right)}}{{1 - {a^2}{t^3}}} \\
\end{array}$
• Mar 9th 2010, 08:12 AM
dbakeg00
Quote:

Originally Posted by Quacky
http://www.mathhelpforum.com/math-he...a638e2e5-1.gif

Yuck.(Surprised) You can certainly take out a factor of a.

I did think about factoring out an "a" but both terms in the denominator don't contain an "a", so I couldn't cancel out anything, right?
• Mar 9th 2010, 08:40 AM
dbakeg00
Quote:

Originally Posted by hameerabbasi
Rationalize before differentiating:

$\begin{array}{l}
\theta = \frac{{1 - a\sqrt[3]{{{t^2}}}}}{{1 + a\sqrt {{t^3}} }} \cdot \frac{{1 - a\sqrt {{t^3}} }}{{1 - a\sqrt {{t^3}} }} \\
= \frac{{\left( {1 - a\sqrt[3]{{{t^2}}}} \right)\left( {1 - a\sqrt {{t^3}} } \right)}}{{1 - {a^2}{t^3}}} \\
\end{array}$

What benefit do you see from rationalizing it first, it still ends up a pretty sloppy mess, unless I'm missing something?
• Mar 9th 2010, 11:08 AM
dbakeg00
What I'm really after is how do you get from here:

$

\frac{(1+at^{\frac{3}{2}})(\frac{-2}{3}at^{\frac{-1}{3}}) - (1-at^{\frac{2}{3}})(\frac{3}{2}at^{\frac{1}{2}})}{(1 +at^{\frac{3}{2}})^2}
$

to here:

$\frac{5a^2\sqrt[6]{t^7} - \frac{4a}{\sqrt[3]{t}} - 9a\sqrt{t}}{6(1+a\sqrt{t^3})^2}$

any help will be very much appreciated. Thanks!
• Mar 9th 2010, 11:13 AM
icemanfan
Multiply out the numerator, combine like terms, take out the common factor of 1/6 by putting 6 in the denominator.
• Mar 9th 2010, 11:16 AM
dbakeg00
Quote:

Originally Posted by icemanfan
Multiply out the numerator, combine like terms, take out the common factor of 1/6 by putting 6 in the denominator.

I did multiply out the numerator...look above. That is where I am stuck. I know the process, I was just wondering if someone could help "baby-step" through this one?
• Mar 9th 2010, 11:32 AM
icemanfan
Quote:

Originally Posted by dbakeg00
I did multiply out the numerator...look above. That is where I am stuck. I know the process, I was just wondering if someone could help "baby-step" through this one?

You made an error when multiplying out the numerator.

$\frac{-2}{3}at^\frac{-1}{3} + \frac{-2}{3}a^2t^{\frac{7}{6}} - \frac{3}{2}at^{\frac{1}{2}} - \frac{3}{2}a^2t^{\frac{7}{6}}$

should be
$\frac{-2}{3}at^\frac{-1}{3} + \frac{-2}{3}a^2t^{\frac{7}{6}} - \frac{3}{2}at^{\frac{1}{2}} + \frac{3}{2}a^2t^{\frac{7}{6}}$.

Then you can combine to get
$\left(\frac{3}{2} - \frac{2}{3}\right)a^2t^{\frac{7}{6}} - \frac{2}{3}at^\frac{-1}{3} - \frac{3}{2}at^{\frac{1}{2}}$.

Simplifying and using a common denominator of 6:
$\frac{5}{6}a^2t^{\frac{7}{6}} - \frac{4}{6}at^\frac{-1}{3} - \frac{9}{6}at^{\frac{1}{2}}$.

Putting the factor of 6 in the denominator leaves you with
$5a^2t^{\frac{7}{6}} - 4at^\frac{-1}{3} - 9at^{\frac{1}{2}}$ in the numerator.
• Mar 9th 2010, 11:41 AM
dbakeg00
Thank you so much. That algebra error I made above was driving me nuts bc everything was not working out like I knew it should. Thank again for taking the time to help me.