Let $\mathbf c=(c_1,c_2,c_3)$ be a constant vector, $\mathbf r=(x,y,z)$ the position vector and $r=||\mathbf r||$. Prove the following: $\mathrm{grad} [(\mathbf c \cdot \mathbf r) r^\alpha]= r^\alpha (1+\alpha) \mathbf c.$
Using product rule, one can simplify
$$\mathrm{grad} [(\mathbf c \cdot \mathbf r) r^\alpha]=r^\alpha\,\color{red}{\mathrm{grad} (\mathbf c \cdot \mathbf r)}+ (\mathbf c \cdot \mathbf r)\,\color{blue}{\mathrm{grad\,} r^\alpha}=r^\alpha \,\color{red}{\mathbf c}+(\mathbf c \cdot \mathbf r) \color{blue}{\alpha r^{\alpha-2}\mathbf r},$$
as $\color{red}{\mathrm{grad} (\mathbf c \cdot \mathbf r)=\mathbf c}$ and $\color{blue}{\mathrm{grad\,} r^\alpha=\alpha r^{\alpha-1}\,\mathrm{grad\,}r=\alpha r^{\alpha-1}\,\frac{\mathbf r}{r}=\alpha r^{\alpha-2}\mathbf r}$.
Now, my question is: Is it possible to simplify $$\color{green}{(\mathbf c \cdot \mathbf r)\alpha r^{\alpha-2}\mathbf r=\alpha r^\alpha \mathbf c \,}?$$ Because if yes, I could add both terms to
$$\mathrm{grad} [(\mathbf c \cdot \mathbf r) r^\alpha]= r^\alpha \,\mathbf c+\color{green}{(\mathbf c \cdot \mathbf r) \alpha r^{\alpha-2}\mathbf r}= r^\alpha \,\mathbf c+ \color{green}{\alpha r^\alpha \mathbf c} = r^\alpha (1+\alpha) \mathbf c$$ which I'm told should be the correct result.
Is the green equality correct? Thanks a lot!
Your green inequality doesn't hold in general. For instance, if $\mathbf c=(1,0,0)$, then the LHS can have non-zero $y$ component.
If the original identity you wanted to prove was correct, it would predict that the directional derivative of $\mathbf c \cdot \mathbf r r^\alpha $ in some direction $\mathbf c^\perp$ with $\mathbf c^\perp\cdot\mathbf c=0$ is $\nabla (\mathbf c \cdot \mathbf r r^\alpha) \cdot \mathbf c^\perp = (...) \mathbf c \cdot \mathbf c^\perp = 0$. Taking $\mathbf c=(1,0,0)$ again to simplify things and $\mathbf c^\perp := (0,1,0)$, we directly compute the directional derivative, i.e. the partial in $y$-
$$ \partial_y (\mathbf c \cdot \mathbf r r^\alpha) = \partial_y (x (x^2+y^2+z^2)^{\alpha/2}) = \frac{\alpha x}2 (x^2+y^2+z^2)^{\alpha/2-1} 2y = \alpha x y (x^2+y^2+z^2)^{\alpha/2-1} = \alpha (\mathbf c \cdot \mathbf r)(\mathbf c^\perp \cdot \mathbf r) r^{\alpha-2}. $$ This is not identically zero (in particular its plainly not zero at the point $\mathbf r = \mathbf c + \mathbf c^\perp$). So your question isn't correct.