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- Thread starter daudaudaudau
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- #2

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then ##\nabla f(A) = \hat x \frac{\partial f}{\partial x}\left[\frac{\partial a}{\partial x}+\frac{\partial b}{\partial x}+\frac{\partial c}{\partial x} \right] +

\hat y\frac{\partial f}{\partial y}\left[\frac{\partial a}{\partial y}+\frac{\partial b}{\partial y}+\frac{\partial c}{\partial y} \right]+

\hat z \frac{\partial f}{\partial z}\left[\frac{\partial a}{\partial z}+\frac{\partial b}{\partial z}+\frac{\partial c}{\partial z} \right]##

##\nabla f = \hat x \frac{\partial f}{\partial x}+ \hat y \frac{\partial f}{\partial y}+ \hat z \frac{\partial f}{\partial z}##

Also, ## \nabla A = \pmatrix{\frac{\partial a}{\partial x} &\frac{\partial b}{\partial x} & \frac{\partial c}{\partial x}\\

\frac{\partial a}{\partial y} &\frac{\partial b}{\partial y} & \frac{\partial c}{\partial y} \\

\frac{\partial a}{\partial z} &\frac{\partial b}{\partial z} & \frac{\partial c}{\partial z}}##

So if you carry out the matrix math and rearrange the terms,

You will see that the equation is true.

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Thanks. How would that look in spherical coordinates ?

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Note that the ##i##th component of ##\nabla(f\circ A)(x)## is ##(f\circ A)_{,i}(x)##.

$$(f\circ A)_{,i}(x)= f_{,j}(A(x))A^j{}_{,i}(x) =\nabla f(A(x))\cdot A_{,i}(x) =(\nabla f\circ A)(x)\cdot A_{,i}(x).$$ I suppose we could also write this as

$$\nabla(f\circ A)(x) =(\nabla f\circ A)(x)\cdot\nabla A(x),$$ but I don't see why we'd want to.

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