September 7th, 2017, 08:57 AM  #1 
Newbie Joined: Sep 2017 From: Latvia/Denmark Posts: 13 Thanks: 0  Differentiation Hey, I don't quite understand why isn't the answer just 3x squared +2, but also has a part under the line? Last edited by skipjack; September 7th, 2017 at 09:21 AM. 
September 7th, 2017, 09:37 AM  #2 
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0 
I don't know how much you know about derivation, but: sqrt (some function) ' = 1/2sqrt(some function) * (the function)' sqrt(x)' = 1/2*sqrt(x) * (x)' ( from the table) . Here our function is (x) so x' is one so we are left with 1/2*sqrt(x) so: (x^3+2x+1)^(1/2) is actually sqrt of (x^3+2x+1) so following the table: (sqrt(x^3+2x+1))' = 1/2*sqrt(x^3+2x+1) buuut you also have to take the first derivative of the function in the root. Here the function is x^3+2x+1, derivative: 3x^2+2 Finally: 3x^2+2/2sqrt(x^3+2x+1). If I was not clear enough, tell me... I'll try to explain better. Last edited by skipjack; September 7th, 2017 at 09:39 AM. 
September 7th, 2017, 09:38 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1385 
View the same video from time index 10:00. It explains why the derivative is obtained as a product. One factor has an exponent of 1/2, which is equivalent to using a denominator containing the same factor with its exponent changed to 1/2.

September 7th, 2017, 09:41 AM  #4  
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  Quote:
 
September 7th, 2017, 09:49 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1385 
You omitted the prime from the LHS of the second line in the image.

September 8th, 2017, 02:56 AM  #6 
Newbie Joined: Sep 2017 From: Latvia/Denmark Posts: 13 Thanks: 0 
I understood when you explained, but how can I know when to use which formula?

September 8th, 2017, 03:28 AM  #7  
Member Joined: May 2017 From: Slovenia Posts: 87 Thanks: 0  Quote:
I found one article I think it will help you understand: http://www.math.wustl.edu/~freiwald/...ativetable.pdf  

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