Can I extend chain rule for gradients like this?

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The chain rule for $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R}^n \to \mathbb{R}$ is

$$\nabla (f \circ g) = \left( f' \circ g \right) \cdot \nabla (g)$$

so, can I write, $\nabla (f'(r)) = f''(r) \nabla (r)$ ?

(since we can prove that $grad (f(r)) = f'(r) grad (r)$ I think that the above result should be true as well)