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December 15th, 2012, 04:10 AM  #1 
Newbie Joined: Dec 2012 Posts: 8 Thanks: 0  First step of derivative, and also second derviative
Hi Everyone, Quick question for most any math expert. If asked to find the first derivative of: f(x)=acos2x+b(sinx)^2 apparently the first step is f ' (x) = 2asin(2x) + 2bsinxcosx I do know that the derivative of cosx=sinx and that the derivative of sinx=cosx. But I guess we dont use the product rule for the first term and then again for the second term. And I also have no idea where he got the 2 in the first term NOR do I understand where the cosx comes from in the second term. Can anyone offer a why? Also, regarding the second derivative, how does it go from f ' (x)= (b2a)sin2x to f '' (x) = (b2a)cox2X * 2 Any help is GREATLY appreciated! 
December 15th, 2012, 10:05 AM  #2  
Senior Member Joined: May 2011 Posts: 501 Thanks: 6  Re: First step of derivative, and also second derviative
Hi Everyone, Quick question for most any math expert. If asked to find the first derivative of: f(x)=acos2x+b(sinx)^2 apparently the first step is f ' (x) = 2asin(2x) + 2bsinxcosx Quote:
The chain rule basically says, derivative of outside times derivative of inside. So, we get Quote:
Again, the chain rule. Just like the derivative of is . The derivative of the 'outside' is . The derivative of is So, we get Quote:
In other words, the chain rule says, Let and write So, and So, we have Any help is GREATLY appreciated!  

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