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Looking at some of the definitions of gradient (for a scalar function $f(\mathbf{r}(u)))$; $$\frac{\mathrm{d}f}{\mathrm{d}\mathbf{r}} = \nabla f $$ (from http://onlinelibrary.wiley.com/doi/10.1002/0471705195.app3/pdf) $$\mathrm{d}f = \nabla f \cdot \mathrm{d}\mathbf{r}$$ (usual definition)

Now if I look at the Chain rule, we know; $$\frac{\mathrm{d}f}{du} = \nabla f \cdot \frac{\mathrm{d}\mathbf{r}}{du}$$

Now from here on is my own working; $$\frac{\mathrm{d}f}{du} = \frac{\mathrm{d}f}{d\mathbf{r}} \frac{\mathrm{d}\mathbf{r}}{du} = \nabla f \frac{\mathrm{d}\mathbf{r}}{du}$$ Which is not the same as above.

Can someone explain what I have done that is wrong, and why it is wrong?

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    When you say $\frac{\mathrm{d}f}{du} = \dots= \nabla f \frac{\mathrm{d}\mathbf{r}}{du}$, what exactly do you mean by the multiplication on the right? – 2017-02-06
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    On the left, df/du is a scalar value. On the right the product of two vectors. That product *must* be the dot product of the two vectors. – 2017-09-22

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