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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)= (b-2a)sin2x to f '' (x) = (b-2a)cox2X * 2 Any help is GREATLY appreciated!
December 15th, 2012, 10:05 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 $-2a\sin (2x)$

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

Again, the chain rule. Just like the derivative of $x^{2}$ is $2x$.

The derivative of the 'outside' is $2\sin(x)$. The derivative of $\sin(x)$ is $\cos(x)$

So, we get $2\sin(x)\cos(x)$

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, $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

$f'(x)=(b-2a)\sin(2x)$

Let $u=2x$ and write $y=\sin(u)$

So, $\frac{dy}{du}=\cos(u)$

and $\frac{du}{dx}=2$

So, we have $2\cos(2x)$

Any help is GREATLY appreciated!

 Tags derivative, derviative, step

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