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

Chesley 2021-10-03 Answered
Find derivatives of \(\displaystyle{r}={\frac{{{12}}}{{{0}}}}-{\frac{{{4}}}{{{0}^{{{3}}}}}}+{\frac{{{1}}}{{{0}^{{{4}}}}}}\)

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Expert Answer

d2saint0
Answered 2021-10-04 Author has 18562 answers
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,
\(\displaystyle{\frac{{{d}{r}}}{{{d}{0}}}}=-{\frac{{{12}}}{{{0}^{{{2}}}}}}+{\frac{{{12}}}{{{0}^{{{4}}}}}}-{\frac{{{4}}}{{{0}^{{{5}}}}}}\)
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