Math I Can Do

By
Erica
on December 14, 2005 7:27 PM | | Comments (4) | TrackBacks (0)

Someone back on this post requested proof that the derivative of cos(x) is -sin(x). I may not have passed Calc III, this time around, but Calc I is eminently doable. To avoid spaminess here, find the proof here. (Now with Maple goodness.)

Categories:

0 TrackBacks

Listed below are links to blogs that reference this entry: Math I Can Do.

TrackBack URL for this entry: http://www.sperari.com/cgi-bin/mt/mt-tb.cgi/245

4 Comments

By moof on December 15, 2005 2:26 AM

How do you justify lim h-> 0 [ sin(h) / h] ?

By Erica on December 15, 2005 5:47 AM

That part I lazily let Maple do, but it's also doable by L'Hopital's rule since it's the limit of one function that's 0 over another that's 0. The derivative with respect to h of sin(h) is cos(h), which is 1 for h=0; the derivative of h with respect to h is 1, so it's 1/1 = 1.

By moof on December 15, 2005 6:29 AM

but, I mean, isn't using l'hopital a little circular? using the derivative of cos to prove the derivative of sin?

By Erica on December 15, 2005 11:47 AM

Hmm. Most of the similar proofs I've seen online use

lim h-> 0 [sin(h)/h]

as a given, and proceed to use that fact to prove the opposite - that d/dx [sin(x)] is cos(x). I suppose I don't find it any more suspect than using the cos(x+h) identity - I mean, how was that identity discovered in the first place?

Leave a comment

Search

About this Entry

This page contains a single entry by Erica published on December 14, 2005 7:27 PM.

Semester was the previous entry in this blog.

Dear Department Stores, I Hate You, KTHXBAI. is the next entry in this blog.

Find recent content on the main index or look in the archives to find all content.