# Differentiate this expression with respect to x

1. Apr 15, 2007

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

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

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

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