# Find derivatives of r=\frac{12}{0}-\frac{4}{0^{3}}+\frac{1}{0^{4}}

Find derivatives of $$\displaystyle{r}={\frac{{{12}}}{{{0}}}}-{\frac{{{4}}}{{{0}^{{{3}}}}}}+{\frac{{{1}}}{{{0}^{{{4}}}}}}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

d2saint0
Step 1
Given:
$$\displaystyle{r}={\frac{{{12}}}{{{0}}}}-{\frac{{{4}}}{{{0}^{{{3}}}}}}+{\frac{{{1}}}{{{0}^{{{4}}}}}}$$
We use,
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{1}}}{{{x}^{{{n}}}}}}\right)}={\frac{{-{n}}}{{{x}^{{{n}+{1}}}}}}$$
Now differentiate with respect to θ we get,
$$\displaystyle{\frac{{{d}{r}}}{{{d}{0}}}}={\frac{{{d}}}{{{d}{0}}}}{\left({\frac{{{12}}}{{{0}}}}\right)}-{\frac{{{d}}}{{{d}{0}}}}{\left({\frac{{{4}}}{{{0}^{{{3}}}}}}\right)}+{\frac{{{d}}}{{{d}{0}}}}{\left({\frac{{{1}}}{{{0}^{{{4}}}}}}\right)}$$
$$\displaystyle{\frac{{{d}{r}}}{{{d}{0}}}}={12}{\frac{{{d}}}{{{d}{0}}}}{\left({\frac{{{1}}}{{{0}}}}\right)}-{4}{\frac{{{d}}}{{{d}{0}}}}{\left({\frac{{{1}}}{{{0}^{{{3}}}}}}\right)}+{\frac{{{d}}}{{{d}{0}}}}{\left({\frac{{{1}}}{{{0}^{{{4}}}}}}\right)}$$
$$\displaystyle{\frac{{{d}{r}}}{{{d}{0}}}}={12}{\left(-{\frac{{{1}}}{{{0}^{{{2}}}}}}\right)}-{4}{\left({\frac{{-{3}}}{{{0}^{{{4}}}}}}\right)}+{\left({\frac{{-{4}}}{{{0}^{{{5}}}}}}\right)}$$
$$\displaystyle{\frac{{{d}{r}}}{{{d}{0}}}}=-{\frac{{{12}}}{{{0}^{{{2}}}}}}+{\frac{{{12}}}{{{0}^{{{4}}}}}}-{\frac{{{4}}}{{{0}^{{{5}}}}}}$$
Step 2
Therefore,
$$\displaystyle{\frac{{{d}{r}}}{{{d}{0}}}}=-{\frac{{{12}}}{{{0}^{{{2}}}}}}+{\frac{{{12}}}{{{0}^{{{4}}}}}}-{\frac{{{4}}}{{{0}^{{{5}}}}}}$$