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December 15th, 2012, 05:10 AM   #1
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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)= (b-2a)sin2x

to

f '' (x) = (b-2a)cox2X * 2

Any help is GREATLY appreciated!
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December 15th, 2012, 11:05 AM   #2
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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:
And I also have no idea where he got the -2 in the first term
The chain rule. The derivative of cos is -sin. The derivative of 2x is 2.

The chain rule basically says, derivative of outside times derivative of inside.

So, we get


Quote:
NOR do I understand where the cosx comes from in the second term
.

Again, the chain rule. Just like the derivative of is .

The derivative of the 'outside' is . The derivative of is

So, we get

Quote:
Also, regarding the second derivative, how does it go from

f ' (x)= (b-2a)sin2x

to

f '' (x) = (b-2a)cox2X * 2
The chain rule.

In other words, the chain rule says,



Let and write

So,

and

So, we have

Any help is GREATLY appreciated!
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