# Thread: Simplification of a derivative

1. ## 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:

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

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

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

and then this:

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

and here is what I multiplied it all out to:

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

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.

2. Yuck. You can certainly take out a factor of a.

3. ## Rationalize

Rationalize before differentiating:

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

4. 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?

5. Originally Posted by hameerabbasi
Rationalize before differentiating:

$\displaystyle \begin{array}{l} \theta = \frac{{1 - a\sqrt[3]{{{t^2}}}}}{{1 + a\sqrt {{t^3}} }} \cdot \frac{{1 - a\sqrt {{t^3}} }}{{1 - a\sqrt {{t^3}} }} \\ = \frac{{\left( {1 - a\sqrt[3]{{{t^2}}}} \right)\left( {1 - a\sqrt {{t^3}} } \right)}}{{1 - {a^2}{t^3}}} \\ \end{array}$
What benefit do you see from rationalizing it first, it still ends up a pretty sloppy mess, unless I'm missing something?

6. What I'm really after is how do you get from here:

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

to here:

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

any help will be very much appreciated. Thanks!

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

8. 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?

9. 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.

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

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

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

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

Putting the factor of 6 in the denominator leaves you with
$\displaystyle 5a^2t^{\frac{7}{6}} - 4at^\frac{-1}{3} - 9at^{\frac{1}{2}}$ in the numerator.

10. 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.