# How it is equal to (del f)/(del x′)=(del f)/(del x)cos theta+(del f)/(del y)sin theta Can you explain?

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?
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ignaciopastorp6
x and y are functions that depend on x′ and y′, i.e.,
$x\left({x}^{\prime },{y}^{\prime }\right)={x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta ,\phantom{\rule{0ex}{0ex}}y\left({x}^{\prime },{y}^{\prime }\right)={x}^{\prime }\mathrm{sin}\theta +{y}^{\prime }\mathrm{cos}\theta .$
So
$\begin{array}{rl}\frac{\mathrm{\partial }x}{\mathrm{\partial }{x}^{\prime }}& =\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^{\prime }}\left({x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta \right)=\frac{\mathrm{\partial }}{\mathrm{\partial }{x}^{\prime }}\left({x}^{\prime }\mathrm{cos}\theta \right)\\ & =\mathrm{cos}\theta .\end{array}$
Similarly, $\frac{\mathrm{\partial }y}{\mathrm{\partial }{x}^{\prime }}=\mathrm{sin}\theta .$