# Problem 2-9: Differentiating Polynomial and Rational Functions

Differentiate
a) $f\left(x\right)= 1+ 2x$
b) $f\left(x\right)= 1 - 2x + 2x^2$
c) $f\left(x\right)= 1- x^2 +\frac{1}{x}$
d) $f\left(x\right)= 1 - x +\frac{1}{1-x}$

Solution
The following rules can be used:
I) $\displaystyle\frac{d}{dx}ax^n=anx^{n-1}$
II) $\displaystyle\frac{d}{dx}\left(u+v\right)=\frac{du}{dx}+\frac{dv}{dx}$
III) $\displaystyle\frac{d}{dx}\frac{u}{v}=\frac{1}{v^2}\left(v\frac{du}{dx}-u\frac{dv}{dx}\right)$

a) $f\left(x\right)= 1+ 2x$
$\displaystyle\frac{d}{dx}f\left(x\right)= \displaystyle\frac{d\left(1+2x\right)}{dx}=0+2=2$

b) $f\left(x\right)= 1 - 2x + 2x^2$
$\displaystyle\frac{d}{dx}f\left(x\right)= \displaystyle\frac{d\left(1 - 2x + 2x^2\right)}{dx}=\displaystyle\frac{d}{dx}1 + \displaystyle\frac{d}{dx}\left(- 2x\right) + \displaystyle\frac{d}{dx}\left(2x^2\right)=0-2+2\times 2 \times x^{2-1}=4x-2$

c) $f\left(x\right)= 1- x^2 +\displaystyle\frac{1}{x}$
$\displaystyle\frac{d}{dx}f\left(x\right)= \displaystyle\frac{d}{dx} \left(1- x^2 +\displaystyle\frac{1}{x}\right)=\displaystyle\frac{d}{dx} 1+ \displaystyle\frac{d}{dx} \left(- x^2\right)+\displaystyle\frac{d}{dx}\left(\displaystyle\displaystyle\frac{1}{x}\right) =0 -2x^{2-1}+\displaystyle\frac{d}{dx}x^{-1}=-2x +\left(-1\right) \times x^{-1-1}=-2x-\displaystyle\frac{1}{x^2}$

d) $f\left(x\right)= 1 - x +\displaystyle\frac{1}{1-x}$
$\displaystyle\frac{d}{dx}f\left(x\right)= \displaystyle\frac{d}{dx} \left(1 - x +\displaystyle\frac{1}{1-x}\right)= \displaystyle \frac{d}{dx} 1 + \displaystyle\frac{d}{dx} \left(- x\right) + \displaystyle\frac{d}{dx}\displaystyle\frac{1}{1-x}$

Let's use rule III to find $\frac{d}{dx}\frac{1}{1-x}$ .
$u=1, v=x-1, du =0, dv=dx \to \displaystyle\frac{d}{dx}\left(\displaystyle\frac{1}{x}\right)=\displaystyle\frac{d}{dx}\displaystyle\frac{u}{v}=\displaystyle\frac{1}{v^2}\left(v\displaystyle\frac{du}{dx}-u\displaystyle\frac{dv}{dx}\right)$
$=\displaystyle\frac{1}{\left(x-1\right)^2}\left(\left(x-1\right)\displaystyle\frac{du}{dx}-1\times \displaystyle\frac{dv}{dx}\right) = \displaystyle\frac{1}{\left(x-1\right)^2}\left(0-1\right)=\displaystyle\frac{1}{\left(x-1\right)^2}$

Therefore,
$\displaystyle\frac{d}{dx}f\left(x\right)=\displaystyle\frac{d}{dx} 1 + \displaystyle\frac{d}{dx} \left(- x\right) + \displaystyle\frac{d}{dx}\displaystyle\frac{1}{1-x}=0-1+\displaystyle\frac{1}{\left(x-1\right)^2}=\displaystyle\frac{1}{\left(x-1\right)^2}-1$

## 3 thoughts on “Problem 2-9: Differentiating Polynomial and Rational Functions”

1. It's very useful wapsite for us.

2. Fine way of telling, and good article to get data on the topic of my presentation focus, which i am going to convey in university.
{"extraction_info": {"found_math": true, "script_math_tex": 19, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6438513994216919, "perplexity": 6174.327340403193}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783392099.27/warc/CC-MAIN-20160624154952-00051-ip-10-164-35-72.ec2.internal.warc.gz"}