- #1

sams

Gold Member

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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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