- #1
sams
Gold Member
- 84
- 2
I have a question regarding the dot product and the cross product differentiation. I was wondering whether:
$$\frac{d(\vec{A}.\vec{B})}{du} = \vec{A}. \frac{d\vec{B}}{du} + \frac{d\vec{A}}{du} .\vec{B}$$
is the same as
$$\frac{d(\vec{A}.\vec{B})}{du} = \frac{d\vec{A}}{du} .\vec{B} + \vec{A}. \frac{d\vec{B}}{du}$$
and
$$\frac{d(\vec{A}×\vec{B})}{du} = \vec{A}× \frac{d\vec{B}}{du} + \frac{d\vec{A}}{du} ×\vec{B}$$
is the same as
$$\frac{d(\vec{A}×\vec{B})}{du} = \frac{d\vec{A}}{du} ×\vec{B} + \vec{A}× \frac{d\vec{B}}{du}$$
or not!
Thank you so much for your help...
$$\frac{d(\vec{A}.\vec{B})}{du} = \vec{A}. \frac{d\vec{B}}{du} + \frac{d\vec{A}}{du} .\vec{B}$$
is the same as
$$\frac{d(\vec{A}.\vec{B})}{du} = \frac{d\vec{A}}{du} .\vec{B} + \vec{A}. \frac{d\vec{B}}{du}$$
and
$$\frac{d(\vec{A}×\vec{B})}{du} = \vec{A}× \frac{d\vec{B}}{du} + \frac{d\vec{A}}{du} ×\vec{B}$$
is the same as
$$\frac{d(\vec{A}×\vec{B})}{du} = \frac{d\vec{A}}{du} ×\vec{B} + \vec{A}× \frac{d\vec{B}}{du}$$
or not!
Thank you so much for your help...
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