# prove derivative of dot product of A(u) and B(u) with respect to u.

Suppose $$\vec{A}$$ and $$\vec{B}$$ are differentiable functions of a scalar u.
Prove:

$$\Large \frac{d}{du}(\vec{A} \cdot \vec{B}) = \vec{A} \cdot \frac{d\vec{B}}{du} + \frac{d\vec{A}}{du} \cdot \vec{B}$$

$$\Large \frac{df(x)}{dx} = \lim \limits_{\Delta x \to \infty} \frac{f(x+\Delta x) - f(x)}{\Delta x}$$

and apply it to the dot product of $$\vec{A}$$ and $$\vec{B}$$:

$$\Large \frac{d}{du} (\vec{A} \cdot \vec{B}) = \lim \limits_{\Delta u \to \infty} \frac{(\vec{A}+\Delta\vec{A}) \cdot (\vec{B}+\Delta\vec{B}) -\vec{A}\cdot\vec{B}}{\Delta u}$$

appying "foil" method to expand dot produt in numerator:

$$\Large \frac{d}{du} (\vec{A} \cdot \vec{B}) = \lim \limits_{\Delta u \to \infty} \frac{\vec{A}\cdot\Delta B ~+~ \Delta \vec{A} \cdot \vec{B} ~+~ \Delta A \cdot \Delta B}{\Delta u}$$

$$\Large \frac{d}{du} (\vec{A} \cdot \vec{B}) = \lim \limits_{\Delta u \to \infty} \frac{\vec{A}\cdot\Delta B}{\Delta u} + \lim \limits_{\Delta u \to \infty} \frac{\Delta \vec{A} \cdot \vec{B}}{\Delta u} + \lim \limits_{\Delta u \to \infty} \frac{\Delta A \cdot \Delta B}{\Delta u}$$

$$\Large \frac{d}{du} (\vec{A} \cdot \vec{B}) = \vec{A} \cdot \frac{d\vec{B}}{du} + \frac{d\vec{A}}{du} \cdot \vec{B} + \lim \limits_{\Delta u \to \infty} \frac{\Delta A \cdot \Delta B}{\Delta u}$$

now for the part that I don't understand..why does this piece goto zero:

$$\Large \lim \limits_{\Delta u \to \infty} \frac{\Delta A \cdot \Delta B}{\Delta u} = 0?$$

• In your case, $f(x + \Delta x)$ corresponds to $(A \cdot B)(u + \Delta u)$. – Viktor Glombik Nov 20 at 23:26
• @ViktorGlombik Yes. But Leaky's using the shorthand $A+\Delta A$ for $A(u)+A(\Delta u)$ – Graham Kemp Nov 20 at 23:49

Multiplying top and bottom by $$\Delta u$$, we get \begin{align} \lim_{\Delta u \rightarrow 0}\frac {\Delta\vec A\cdot\Delta\vec B}{\Delta u} &=\lim_{\Delta u \rightarrow 0}\frac {\Delta\vec A}{\Delta u} \cdot\frac{\Delta\vec B}{\Delta u}~\Delta u \\[1ex]&= \frac {d\vec A}{du}\cdot \frac {d\vec B}{du}\lim_{\Delta u \rightarrow 0} \Delta u\\[1ex] &= 0\end{align}
I assume we are working in $$\mathbb{R}^n$$, take $$\{e_i\}$$ the standard orthonormal basis.
Then, $$\mathbf{A}(u) = \sum a_i(u) e_i$$, $$\mathbf{B}(u) = \sum b_i(u) e_i$$, and $$(\mathbf{A}\cdot\mathbf{B})(u) = \sum a_i(u) b_i(u)$$. Now evaluating the derivative is much easier because we only have to evaluate each term in the sum. We do that with the product rule:
$$$$\frac{d}{du}(\mathbf{A}\cdot\mathbf{B} )(u) = \sum a_i(\frac{d}{du}b_i) + \sum b_i(\frac{d}{du}a_i) = \mathbf{A}\cdot\frac{d\mathbf{B}}{du} + \frac{d\mathbf{A}}{du} \cdot\mathbf{B}.$$$$