Find the derivatives of the functions r=120−403+104

Step 1
The rate of change in value of dependent variable with respect to the change in value of an independent variable is known as the derivative of the function. According to the quotient rule of differentiation,
${\displaystyle \frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{f^{\prime }\left(x\right)g\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}}$
Also, the differentiation of the expression of the form ${\displaystyle x^{n}isn{x}^{n-1}}$. So, we need to use the quotient rule of differentiation to find the derivative of the function.
Step 2
Using the quotient rule of differentiation, differentiate the function ${\displaystyle r=\frac{12}{0}-\frac{4}{0^3}+\frac{1}{0^4}}$ with respect to θ as follows:
${\displaystyle \frac{dr}{d0}=\frac{d}{d0}(\frac{12}{0}-\frac{4}{0^3}+\frac{1}{0^4})}$
${\displaystyle =\frac{d}{d0}\left(\frac{12}{0}\right)-\frac{d}{d0}\left(\frac{4}{0^3}\right)+\frac{d}{d0}\left(\frac{1}{0^4}\right)}$
${\displaystyle =\frac{0-12}{0^2}-\frac{0-4\cdot {30}^2}{0^6}+\frac{0-{40}^3}{0^8}}$
${\displaystyle =-\frac{12}{0^2}+\frac{12}{0^4}-\frac{4}{0^5}}$
Therefore, the derivative of the function is ${\displaystyle \frac{dr}{d0}=-\frac{12}{0^2}+\frac{12}{0^4}-\frac{4}{0^5}}$.