Answered: "Find derivatives of r=120−403+104" - Plainmath

Chesley

Answered question

2021-10-03

Find derivatives of

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

Answer & Explanation

d2saint0

Skilled2021-10-04Added 89 answers

Step 1
Given:

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

We use,

\frac{d}{d x} (\frac{1}{x^{n}}) = \frac{- n}{x^{n + 1}}

Now differentiate with respect to θ we get,

\frac{d r}{d 0} = \frac{d}{d 0} (\frac{12}{0}) - \frac{d}{d 0} (\frac{4}{0^{3}}) + \frac{d}{d 0} (\frac{1}{0^{4}})

\frac{d r}{d 0} = 12 \frac{d}{d 0} (\frac{1}{0}) - 4 \frac{d}{d 0} (\frac{1}{0^{3}}) + \frac{d}{d 0} (\frac{1}{0^{4}})

\frac{d r}{d 0} = 12 (- \frac{1}{0^{2}}) - 4 (\frac{- 3}{0^{4}}) + (\frac{- 4}{0^{5}})

\frac{d r}{d 0} = - \frac{12}{0^{2}} + \frac{12}{0^{4}} - \frac{4}{0^{5}}

Step 2
Therefore,

\frac{d r}{d 0} = - \frac{12}{0^{2}} + \frac{12}{0^{4}} - \frac{4}{0^{5}}

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