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

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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Alara Mccarthy

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.
$$\displaystyle\frac{d}{{\left.{d}{x}\right.}}{\left({x}^{2}\right)}={2}{x}:$$
$$\displaystyle={3}{\left({4}{x}^{2}{y} \cos{{\left({4}{x}{y}\right)}}+{2}{x} \sin{{\left({4}{x}{y}\right)}}\right)}$$