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

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

ok so i start with definition of derivative:

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

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:

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

  • 1
    $\begingroup$ In your case, $f(x + \Delta x)$ corresponds to $(A \cdot B)(u + \Delta u)$. $\endgroup$ – Viktor Glombik Nov 20 at 23:26
  • $\begingroup$ @ViktorGlombik Yes. But Leaky's using the shorthand $A+\Delta A$ for $A(u)+A(\Delta u)$ $\endgroup$ – 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:

\begin{equation} \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}. \end{equation}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.