Gaintentavyw4

2022-09-14

There are given equations,

$$x={x}^{\prime}\mathrm{cos}\theta -{y}^{\prime}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}y={x}^{\prime}\mathrm{sin}\theta +{y}^{\prime}\mathrm{cos}\theta $$

$$\frac{\mathrm{\partial}f}{\mathrm{\partial}{x}^{\prime}}=\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\frac{\mathrm{\partial}x}{\mathrm{\partial}{x}^{\prime}}+\frac{\mathrm{\partial}f}{\mathrm{\partial}y}\frac{\mathrm{\partial}y}{\mathrm{\partial}{x}^{\prime}}$$

I don't understand how it is equal to

$$\frac{\mathrm{\partial}f}{\mathrm{\partial}{x}^{\prime}}=\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{cos}\theta +\frac{\mathrm{\partial}f}{\mathrm{\partial}y}\mathrm{sin}\theta $$

Can you explain?

$$x={x}^{\prime}\mathrm{cos}\theta -{y}^{\prime}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}y={x}^{\prime}\mathrm{sin}\theta +{y}^{\prime}\mathrm{cos}\theta $$

$$\frac{\mathrm{\partial}f}{\mathrm{\partial}{x}^{\prime}}=\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\frac{\mathrm{\partial}x}{\mathrm{\partial}{x}^{\prime}}+\frac{\mathrm{\partial}f}{\mathrm{\partial}y}\frac{\mathrm{\partial}y}{\mathrm{\partial}{x}^{\prime}}$$

I don't understand how it is equal to

$$\frac{\mathrm{\partial}f}{\mathrm{\partial}{x}^{\prime}}=\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{cos}\theta +\frac{\mathrm{\partial}f}{\mathrm{\partial}y}\mathrm{sin}\theta $$

Can you explain?

ignaciopastorp6

Beginner2022-09-15Added 14 answers

x and y are functions that depend on x′ and y′, i.e.,

$$x({x}^{\prime},{y}^{\prime})={x}^{\prime}\mathrm{cos}\theta -{y}^{\prime}\mathrm{sin}\theta ,\phantom{\rule{0ex}{0ex}}y({x}^{\prime},{y}^{\prime})={x}^{\prime}\mathrm{sin}\theta +{y}^{\prime}\mathrm{cos}\theta .$$

So

$$\begin{array}{rl}\frac{\mathrm{\partial}x}{\mathrm{\partial}{x}^{\prime}}& =\frac{\mathrm{\partial}}{\mathrm{\partial}{x}^{\prime}}({x}^{\prime}\mathrm{cos}\theta -{y}^{\prime}\mathrm{sin}\theta )=\frac{\mathrm{\partial}}{\mathrm{\partial}{x}^{\prime}}({x}^{\prime}\mathrm{cos}\theta )\\ & =\mathrm{cos}\theta .\end{array}$$

Similarly, $\frac{\mathrm{\partial}y}{\mathrm{\partial}{x}^{\prime}}=\mathrm{sin}\theta .$

$$x({x}^{\prime},{y}^{\prime})={x}^{\prime}\mathrm{cos}\theta -{y}^{\prime}\mathrm{sin}\theta ,\phantom{\rule{0ex}{0ex}}y({x}^{\prime},{y}^{\prime})={x}^{\prime}\mathrm{sin}\theta +{y}^{\prime}\mathrm{cos}\theta .$$

So

$$\begin{array}{rl}\frac{\mathrm{\partial}x}{\mathrm{\partial}{x}^{\prime}}& =\frac{\mathrm{\partial}}{\mathrm{\partial}{x}^{\prime}}({x}^{\prime}\mathrm{cos}\theta -{y}^{\prime}\mathrm{sin}\theta )=\frac{\mathrm{\partial}}{\mathrm{\partial}{x}^{\prime}}({x}^{\prime}\mathrm{cos}\theta )\\ & =\mathrm{cos}\theta .\end{array}$$

Similarly, $\frac{\mathrm{\partial}y}{\mathrm{\partial}{x}^{\prime}}=\mathrm{sin}\theta .$

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