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I understand there are multiple ways of of proving the product rule for the derivative of an inner product, though I cannot figure out how to do this one specifically:

let $\alpha,\beta :R \rightarrow R^n$ be differentiable functions. If $ f(t)=\langle \alpha(t),\beta(t) \rangle,$ using only these three rules (IE the conditions for inner product):

$\langle x,x \rangle > 0$ if $x \not= 0$

$\langle x,y \rangle = \langle y,x \rangle $

$ \langle ax+by,z \rangle = a\langle x,z \rangle + b\langle y,z \rangle $

show that

$f'(t) = \langle \alpha(t), \beta'(t) \rangle + \langle \alpha'(t), \beta(t) \rangle $

Basically I understand how to do it using summations, the normal product rule and and the linearity of summations but I do not understand how to get this result another way. There was another question asked that did it by taking the limits, which also makes sense to me.

thanks for any suggestions/help.

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1 Answer 1

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Use the fact that:

$$ f(t + h) = \langle a(t+h), b(t + h)\rangle \approx \langle a(t) + h a'(t), b(t) + h b'(t)\rangle$$

Use the properties you mentioned above to expand this out. Then subtract away $f(t)$, divide by $h$ and take limits. You'll notice by subtracting $f(t)$, we'll remove all the terms without $h$. By dividing by $h$ and taking limits, we'll drop off the $h^2$ term which is next to the $\langle a'(t), b'(t) \rangle$ term.

You can make the whole thing more precise by taking limits from the start where I simply used $\approx$.

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  • $\begingroup$ So what you are saying is i can assume that $f(t)$ is linear, because f(t+h) - f(t) = f(h) ? $\endgroup$
    – Neo
    Mar 18, 2013 at 2:41
  • $\begingroup$ i think i got it, thanks for the guidance $\endgroup$
    – Neo
    Mar 18, 2013 at 3:21

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