Check whether the given function is satisfying the Fubini’s theorem about ‎mixed partial derivative

ka1leE 2021-09-08 Answered

Check whether the given function is satisfying the Fubini’s theorem about ‎mixed partial derivative ‎
\(\displaystyle f{{\left({x},{y}\right)}}={3}{x}^{2} \sin{{\left({4}{x}{y}\right)}}\)

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Expert Answer

Alara Mccarthy
Answered 2021-09-09 Author has 4221 answers

Possible derivation:
\(\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({3}{x}^{2} \sin{{\left({4}{x}{y}\right)}}\right)}\)
Factor out constants:
\(\displaystyle={3}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2} \sin{{\left({4}{x}{y}\right)}}\right)}\right)}\)
Use the product rule, \(\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({u}{v}\right)}={v}\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}+{u}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}\), where \(\displaystyle{u}={x}^{2}{\quad\text{and}\quad}{v}= \sin{{\left({4}{x}{y}\right)}}:\)
\(\displaystyle={3}{x}^{2}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left( \sin{{\left({4}{x}{y}\right)}}\right)}\right)}+{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}\right)} \sin{{\left({4}{x}{y}\right)}}\)
Using the chain rule, \(\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left( \sin{{\left({4}{x}{y}\right)}}\right)}=\frac{{{d} \sin{{\left({u}\right)}}}}{{{d}{u}}}\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}\), where \(u = 4 x y\) and \(\displaystyle\frac{d}{{{d}{u}}}{\left( \sin{{\left({u}\right)}}\right)}= \cos{{\left({u}\right)}}:\)
\(\displaystyle={3}{\left({\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}\right)} \sin{{\left({4}{x}{y}\right)}}+ \cos{{\left({4}{x}{y}\right)}}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({4}{x}{y}\right)}\right)}{x}^{2}\right)}\)
Factor out constants:
\(\displaystyle={3}{\left({\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}\right)} \sin{{\left({4}{x}{y}\right)}}+{4}{y}{\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}\right)}\right)}{x}^{2} \cos{{\left({4}{x}{y}\right)}}\right)}\)
The derivative of x is 1:
\(\displaystyle={3}{\left({\left(\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}\right)} \sin{{\left({4}{x}{y}\right)}}+{1}{4}{x}^{2}{y} \cos{{\left({4}{x}{y}\right)}}\right)}\)
Use the power rule, \(\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{n}\right)}={n}{x}^{{{n}-{1}}}\), where n = 2.
\(\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}={2}{x}:\)
Answer:
\(\displaystyle={3}{\left({4}{x}^{2}{y} \cos{{\left({4}{x}{y}\right)}}+{2}{x} \sin{{\left({4}{x}{y}\right)}}\right)}\)

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