How can I easily simplify a derivative of sin^-1 (2x + 1)?

[communitycoll]

Homework Statement

Find derivative of sin^-1 (2x + 1)

Homework Equations

I do everything Wolfram does except I don't know how to simplify to get the final answer.

http://www.wolframalpha.com/input/?i=derivative+sin^-1+(2x+++1)

The Attempt at a Solution

Everything Wolfram does basically.

 

[HallsofIvy]

I don't see how you can expect anyone to tell you how to simplify what you got to Wolfram's formula, if you don't tell us what you got!

 

[communitycoll]

Ah terribly sorry. 2 / sqrt(1 - (2x + 1)^2); the last thing in the show steps part for the derivative in Wolfram.

 

[HallsofIvy]

Of course, you know that (2x+1)^2= 4x^2+ 4x+ 1. That means 1- (2x+1)^2= 1 -4x^2- 4x- 1= -4x^2- 4x= -4(x^2- x)= -4x(x-1).

 

[communitycoll]

Of course, you know that (2x+1)^2= 4x^2+ 4x+ 1. That means 1- (2x+1)^2= 1 -4x^2- 4x- 1= -4x^2- 4x= -4(x^2- x)= -4x(x-1).

But if -4x(x-1) is under a radical (as well as being multiplied) how is it that I pull a 2 out of that?

 

[QED Andrew]

But if -4x(x-1) is under a radical (as well as being multiplied) how is it that I pull a 2 out of that?

\sqrt{-4x(x-1)} = \sqrt{4 \cdot -x(x-1)} = \sqrt{4} \sqrt{-x(x-1)} = 2\sqrt{-x(x-1)}

It should be noted 1 - (2x + 1)^2 = -4x(x+1). I'll leave it to you to show \sqrt{-4x(x+1)} = 2\sqrt{-x(x+1)}.

 

[HallsofIvy]

But if -4x(x-1) is under a radical (as well as being multiplied) how is it that I pull a 2 out of that?

What, exactly, do you think the square root of 4 is?

 

[communitycoll]

Ah. I understand now. It took me a bit, but I understand now. Thanks.