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 September 7th, 2017, 08:57 AM #1 Newbie   Joined: Sep 2017 From: Latvia/Denmark Posts: 21 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: 89 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: 18,953 Thanks: 1599 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
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Quote:
 Originally Posted by sarajoveska 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.
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 September 7th, 2017, 09:49 AM #5 Global Moderator   Joined: Dec 2006 Posts: 18,953 Thanks: 1599 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: 21 Thanks: 0 I understood when you explained, but how can I know when to use which formula?
September 8th, 2017, 03:28 AM   #7
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 Originally Posted by vizzy22 I understood when you explained, but how can I know when to use which formula?
Well you have a table and you also have rules about how to find derivative of product and similar. Find those online or in your book. You have to know that table (except in case your teacher gives you those on exam, mine don't). And then every time you see square root you will use this formula, when there is cos or sin you will use the formula for them etc...

http://www.math.wustl.edu/~freiwald/...ativetable.pdf

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