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I have not been able to make the following problem:

Consider that $f$ is a function of only two variables, $y$ and $z$. Show that the gradient:

$$\nabla f=\left(\frac{\partial f}{\partial y}\right)\hat{e_{y}}+\left(\frac{\partial f}{\partial z}\right)\hat{e_{z}}$$

transforms as a vector under rotations.

My idea is to use relationships:

$$\bar{y}=y\cos\phi+z\sin\phi$$ $$\bar{z}=-y\sin\phi+z\cos\phi$$

Solving this system of equations for $y$ and $z$, and determining the derivative: $\partial y/\partial \bar{y}$, $\partial z/\partial \bar{y}$, $\partial y/\partial \bar{z}$ y $\partial z/\partial \bar{y}$.

I can perform these steps without any problem, but then I do not know what to do. If anyone could help me I would appreciate it too.

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Well, given your idea you can then write down the gradient in the new coordinates, using the chain rule and plugging in. For instance, for the $x$ component of $\nabla f$,

$$ \frac{\partial f}{\partial \bar{x}} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \bar{x}} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \bar{x}} $$

On the other hand, the transformation you proposed, can be seen as a matrix-vector operation

$$ \left[ \begin{array}{c} \bar{x}\\ \bar{y} \end{array} \right] = \left[ \begin{array}{cc} \cos\phi & \sin\phi\\ -\sin \phi & \cos\phi \end{array} \right] \left[ \begin{array}{c} x\\ y \end{array} \right] $$

If the two are equal, then you prove your claim.

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