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StevenMx February 2nd, 2009 04:09 PM

Problem with derivative
 
I would be really grateful if anyone could help me with this derivative:



OK, I believe I need to use the quotient rule, and I'll get something like that:



but I don't know what to do next...

StevenMx February 3rd, 2009 03:20 AM

Re: Problem with derivative
 
Updated my previous post :)

CogitoErgoCogitoSum February 3rd, 2009 09:46 AM

Re: Problem with derivative
 
Quote:

Originally Posted by StevenMx

but I don't know what to do next...

First of all, your quotient rule is wrong. You add, not multiply, the two terms


Now, from trigonometry, we ought to know that:




The real question is this... do you know how to find the derivative of an inverse function?

Follow this procedure for any arbitrary inverse function...

Suppose

Then by definition of an inverse function

Find the derivative: utilizing both chain rule and implicit differentiation.

Solve for y':

Plug in your function y in terms of x:


There... thats all you need to know for an arbitrary f(x)... you can find the derivative of any f^-1.

Given that in your problem
,

we know that
and therefore









Refer back to your original problem:


Remembering your chain rules


You have to look back at good old trigonometry yet again...

and so


Plugging in:




This problem is boring me. I will quit now.

StevenMx February 4th, 2009 03:18 AM

Re: Problem with derivative
 
Thanks man :) I really appreciate that.

shynthriir February 4th, 2009 04:58 AM

Re: Problem with derivative
 
Quote:

First of all, your quotient rule is wrong. You add, not multiply, the two terms
Actually, you subtract, not add...

d/dx ( u / v) = (u' * v - u * v') / v^2

CogitoErgoCogitoSum February 4th, 2009 06:56 AM

Re: Problem with derivative
 
Quote:

Originally Posted by shynthriir
Quote:

First of all, your quotient rule is wrong. You add, not multiply, the two terms
Actually, you subtract, not add...

d/dx ( u / v) = (u' * v - u * v') / v^2

LOL. you certainly do. Im sorry man, I totally screwed that up for you. Just change the signs.

StevenMx February 4th, 2009 07:14 AM

Re: Problem with derivative
 
Don't worry man, nothing happened :) I changed the signs.


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