# Chain rule of partial derivatives for composite functions. Function of the form f(x^2+y^2) How do I find the partial derivatives df/dy, df/dx

Chain rule of partial derivatives for composite functions.
Function of the form
$f\left({x}^{2}+{y}^{2}\right)$
How do I find the partial derivatives
$\frac{\mathrm{\partial }f}{\mathrm{\partial }y},\frac{\mathrm{\partial }f}{\mathrm{\partial }x}$
How $f\left({x}^{2}+{y}^{2}\right)$ behaves. Assuming it should of the form
$g\left(x,y\right)\cdot 2y\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}h\left(x,y\right)\cdot 2x$
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ticotaku86
Since $f\left(.\right)$ is univariate function we have
$\frac{\mathrm{\partial }f}{\mathrm{\partial }x}=2x{f}^{\prime }\left({x}^{2}+{y}^{2}\right)$
similarly
$\frac{\mathrm{\partial }f}{\mathrm{\partial }y}=2y{f}^{\prime }\left({x}^{2}+{y}^{2}\right)$
therefore the differential is
$df=2x{f}^{\prime }\left({x}^{2}+{y}^{2}\right)dx+2y{f}^{\prime }\left({x}^{2}+{y}^{2}\right)dy$
therefore
$g\left(x,y\right)=h\left(x,y\right)={f}^{\prime }\left({x}^{2}+{y}^{2}\right)$
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mikioneliir
Use some different notation:
$f\left({x}^{2}+{y}^{2}\right)=:f\left(g\left(x,y\right)\right)=\left(f\circ g\right)\left(x,y\right)$
Now the partial derivatives can be computed by the chain rule (in multiple dimensions):
$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial g}\frac{\partial g}{\partial x}=\frac{\partial f}{\partial g}2x.$
$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial g}\frac{\partial g}{\partial y}=\frac{\partial f}{\partial g}2y.$
Since you don't know anything else about $f$, you can't simplify these terms further.