| 
November 6th, 2009, 10:33 AM
| | Newbie | | Join Date: Nov 2009
Posts: 2
Thanks: 0
Thanked 0 Times in 0 Posts
| |  Integration/ Differentiation
Use the derivative of cosθ to show that d/dθ ( secθ)= secθ tan θ
attempt:
d/dθ( cosθ) = -sinθ
... |
November 6th, 2009, 10:39 AM
| | Senior Member | | Join Date: May 2009 Location: New York
Posts: 415
Country:  Thanks: 10
Thanked 149 Times in 144 Posts
| |
Quote:
Originally Posted by mathcalculator22122  Use the derivative of cosθ to show that d/dθ ( secθ)= secθ tan θ
attempt:
d/dθ( cosθ) = -sinθ
... | let's change  to  
Now use the quotient rule to take the derivative of both sides..
__________________
Using elementary concubinary logic, you can easily show that the flumex is both semi-dependent and, even more importantly, quasi-invariant |
November 6th, 2009, 10:41 AM
| | Newbie | | Join Date: Nov 2009
Posts: 9
Country: Thanks: 0
Thanked 0 Times in 0 Posts
| |
The best way to go about proving that is to use the quotient rule.
If  , then 
Plug in  and  |
November 6th, 2009, 10:47 AM
| | Senior Member | | Join Date: Oct 2008
Posts: 283
Country:  Thanks: 7
Thanked 73 Times in 72 Posts
| |
Just in case a picture helps... 
... where 
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which subject to the chain rule).
__________________________________________
Don't integrate - balloontegrate! Balloon Calculus: Gallery Balloon Calculus Drawing with LaTeX and Asymptote! |
November 6th, 2009, 10:50 AM
| | Senior Member | | Join Date: May 2009 Location: New York
Posts: 415
Country: Thanks: 10
Thanked 149 Times in 144 Posts
| |
that isn't quicker at all, because then you have to bring the powers back down into the denominator to realize you actually have sec and tan
__________________
Using elementary concubinary logic, you can easily show that the flumex is both semi-dependent and, even more importantly, quasi-invariant |
November 6th, 2009, 11:13 AM
| | Senior Member | | Join Date: Oct 2008
Posts: 283
Country: Thanks: 7
Thanked 73 Times in 72 Posts
| |
Prettier though, surely! And how about this?  |
November 6th, 2009, 01:43 PM
| | Senior Member | | Join Date: May 2009 Location: New York
Posts: 415
Country: Thanks: 10
Thanked 149 Times in 144 Posts
| |
I'm not saying it isnt valid or that you won't get the correct answer, but I can apply the chain rule without drawing circles and lines, and as such, it will be quicker
__________________
Using elementary concubinary logic, you can easily show that the flumex is both semi-dependent and, even more importantly, quasi-invariant |
November 6th, 2009, 02:10 PM
| | Senior Member | | Join Date: Oct 2008
Posts: 283
Country: Thanks: 7
Thanked 73 Times in 72 Posts
| |
Quote:
Originally Posted by artvandalay11 I'm not saying it isnt valid or that you won't get the correct answer, but I can apply the chain rule without drawing circles and lines, and as such, it will be quicker | Yes, and it won't create that 'powers' problem, will it?
But now the goal posts have moved, let's not bicker... I may have provoked you in the first place by daring to introduce my pic with 'quicker by balloon!', which I edited out within the minute because I remembered that this is only true for quotients already having a power in the denominator - see http://www.ballooncalculus.org/examp...ence.html#quot if interested - hope so!
Cheers
Tom |
November 6th, 2009, 06:34 PM
| | Newbie | | Join Date: Nov 2009
Posts: 2
Thanks: 0
Thanked 0 Times in 0 Posts
| |
thanks for helping  | | Thread Tools | | | | Display Modes |  Linear Mode | 
Posting Rules
| You may not post new threads You may not post replies You may not post attachments You may not edit your posts
HTML code is Off | | | All times are GMT -7. The time now is 12:59 PM. | | |
