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

Problem 2-9-a
b) $f\left(x\right)= 1 - 2x + 2x^2$
$\displaystyle\frac{d}{dx}f\left(x\right)= \displaystyle\frac{d\left(1 - 2x + 2x^2\right)}{dx}=\displaystyle\frac{d}{dx}1 + \displaystyle\frac{d}{dx}\left(- 2x\right) + \displaystyle\frac{d}{dx}\left(2x^2\right)=0-2+2\times 2 \times x^{2-1}=4x-2$
Problem 2-9-b

c) $f\left(x\right)= 1- x^2 +\displaystyle\frac{1}{x}$
$\displaystyle\frac{d}{dx}f\left(x\right)= \displaystyle\frac{d}{dx} \left(1- x^2 +\displaystyle\frac{1}{x}\right)=\displaystyle\frac{d}{dx} 1+ \displaystyle\frac{d}{dx} \left(- x^2\right)+\displaystyle\frac{d}{dx}\left(\displaystyle\displaystyle\frac{1}{x}\right) =0 -2x^{2-1}+\displaystyle\frac{d}{dx}x^{-1}=-2x +\left(-1\right) \times x^{-1-1}=-2x-\displaystyle\frac{1}{x^2}$
Problem 2-9-c
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,
$\displaystyle\frac{d}{dx}f\left(x\right)=\displaystyle\frac{d}{dx} 1 + \displaystyle\frac{d}{dx} \left(- x\right) + \displaystyle\frac{d}{dx}\displaystyle\frac{1}{1-x}=0-1+\displaystyle\frac{1}{\left(x-1\right)^2}=\displaystyle\frac{1}{\left(x-1\right)^2}-1$
Problem 2-9-d