Differentiate this expression with respect to x

1. Apr 15, 2007

unique_pavadrin

1. The problem statement, all variables and given/known data
Differentiate the following with respect to x leaving in the simplest form
$$\frac{d}{{dx}}\left[ {\ln \left[ {\left( {1 - \sqrt x } \right)^2 \left( {1 + \sqrt x } \right)^3 } \right]} \right]$$

2. Relevant equations
The chain rule
$$\frac{d}{{dx}}\left[ {\left( {f\left( x \right)} \right)^n } \right] = nf\left( x \right)f'\left( x \right)$$
The product rule
$$\frac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)g'\left( x \right) + f'\left( x \right)g\left( x \right)$$
The natural log rule
$$\frac{d}{{dx}}\left[ {\ln f\left( x \right)} \right] = \frac{{f'\left( x \right)}}{{f\left( x \right)}}$$

3. The attempt at a solution

I sort of understand how to differentiate the following, however i cannot get it into the simplest form. What would the best way to differentiate this functions and others which have a similar format?
many thanks for all suggestions and help
unique_pavadrin

2. Apr 15, 2007

cristo

Staff Emeritus
Well use the three rules you quote. Your expression can be written as ln[f(x)g(x)] where $f(x)=(1-\sqrt{x})^2, \hspace{1cm} g(x)=(1+\sqrt{x})^3$. Now, first apply the log rule to give $$\frac{1}{f(x)g(x)}\cdot\frac{d}{dx}[f(x)g(x)]$$. Then apply the product rule, followed by the chain rule. This is the simplest way, as far as I can tell; of course, once you have differentiated, you will need to simplify the expression you obtain.

If you show some work, it'll be easier to help!

[as an aside, this isn't precalculus maths is it?]

Last edited: Apr 15, 2007
3. Apr 15, 2007

chickens

simplify f(x)g(x) using simple algebra properties first then only apply some log rule like log ab = log a + log b
i'm not too sure if d/dx log ab = d/dx log a + d/dx log b, but you may give it a try

4. Apr 15, 2007

unique_pavadrin

thank you for your reply cristo
this problem was given to me in my early stages of calculus, so i though of posting it here, as it what i have considered a fundamental stage, similar to pre-calculus
here is my working for the problem. i first simplified the expression before attempting to differentiate it.
$$\begin{array}{l} \frac{d}{{dx}}\left[ {\ln \left[ {\left( {1 - \sqrt x } \right)^2 \left( {1 + \sqrt x } \right)^3 } \right]} \right] = \frac{d}{{dx}}\left[ {\ln \left[ {\left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 } \right]} \right] \\ \frac{1}{{f\left( x \right)g\left( x \right)}}.\frac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] \\ \frac{d}{{dx}}\left[ {\left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 } \right] = \left( {1 + \sqrt x } \right)\left( {2x} \right) + \left( {x - 1} \right)^2 \left( {\frac{1}{{2\sqrt x }}} \right) \\ \frac{d}{{dx}}\left[ {\left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 } \right] = \frac{{4\sqrt x \left( {1 + \sqrt x } \right) + \left( {x - 1} \right)^2 }}{{2\sqrt x }} \\ \frac{d}{{dx}}\left[ {\ln \left[ {\left( {1 - \sqrt x } \right)^2 \left( {1 + \sqrt x } \right)^3 } \right]} \right] = \frac{{\frac{{4\sqrt x \left( {1 + \sqrt x } \right) + \left( {x - 1} \right)^2 }}{{2\sqrt x }}}}{{\left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 }} = \frac{{4\sqrt x \left( {1 + \sqrt x } \right) + \left( {x - 1} \right)^2 }}{{2\sqrt x \left( {1 + \sqrt x } \right)\left( {x - 1} \right)^2 }} = \frac{{4\sqrt x + 4x + x^2 - 2x + 1}}{{\left( {2\sqrt x + 2x} \right)\left( {x^2 - 2x + 1} \right)}} \\ \end{array}$$

i am a little stuck on how to further simplify this down, as my other working seems to be going around in circles. any help is greatly appreciated.

5. Apr 15, 2007

unique_pavadrin

chickens thank you for your post, i will try that

6. Apr 15, 2007

HallsofIvy

Staff Emeritus
It's always a good idea to simplify as much as possible before differentiating:
$$\ln \left[ {\left( {1 - \sqrt x } \right)^2 \left( {1 + \sqrt x } \right)^3 } \right]= 2 ln(1- \sqrt{x})+ 3ln(1+ \sqrt{x})$$

7. Apr 15, 2007

unique_pavadrin

thanks HallsofIvy

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