Simplification of a derivative

**dbakeg00** · March 9th 2010, 05:43 AM

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.

**Quacky** · March 9th 2010, 05:56 AM

Yuck.

You can certainly take out a factor of a.

**hameerabbasi** · March 9th 2010, 07:08 AM

Rationalize before differentiating:

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

**dbakeg00** · March 9th 2010, 07:12 AM

Originally Posted by Quacky

Yuck.

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?

**dbakeg00** · March 9th 2010, 07:40 AM

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?

**dbakeg00** · March 9th 2010, 10:08 AM

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!

**icemanfan** · March 9th 2010, 10:13 AM

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

**dbakeg00** · March 9th 2010, 10:16 AM

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?

**icemanfan** · March 9th 2010, 10:32 AM

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.

**dbakeg00** · March 9th 2010, 10:41 AM

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.

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