# Quick question on differentiation

1. Aug 11, 2010

### vorcil

1. The problem statement, all variables and given/known data
$$\frac{d}{dx} [ arctan (2x-3) ]$$

3. The attempt at a solution
The general solution for d/dx arctan(x) = $$\frac{du/dx}{1+x^2}$$

so

$$\frac{du/dx}{1+(2x-3)^2}$$

$$du/dx = 2$$

$$\frac{2}{1+(2x-3)^2}$$

$$\frac{2}{1+2x^2-12x+9}$$

$$\frac{2}{10+2x^2-12x}$$

now this is where my question comes in,

When I simplify it, I would divide all the terms in the numerator and denominator by 2,

giving me $$\frac{1}{5+x^2-6x}$$

however i've been told that this simplification is wrong without any explanation

why?

Last edited: Aug 11, 2010
2. Aug 11, 2010

### Dick

It's wrong because (2x-3)^2=4*x^2-12x+9.

3. Aug 11, 2010

### vorcil

Oh right, thanks

also

$$\frac{d}{dx} arcsin(2x-3)$$

=

$$\frac{2}{\sqrt{1-(2x-3)^2}}$$

=

$$\frac{2}{\sqrt{1-(4x^2-12x+9)}}$$

but apparently that's wrong too!
?

it's supposed to be
$$\frac{2}{\sqrt{1-(3-2x)^2}}$$

but i don't understand why,

http://www.wolframalpha.com/input/?i=d/dx+arcsin(2x-3)

4. Aug 11, 2010

### Dick

Those two expressions are exactly the same thing. (2x-3)^2=(3-2x)^2=4x^2-12x+9. Are you overtired?

5. Aug 11, 2010

### vorcil

just one of those days,

6. Aug 11, 2010

### vorcil

Last one I promise,

$$d/dx arcsin(\frac{2x+3}{5})$$

I get from the equation, $$d/dx [ arcsin(x/a) ] = \frac{1}{\sqrt{a^2-x^2}}$$

$$\frac{2}{\sqrt{5^2-(2x+3)^2}}$$

=

$$\frac{2}{\sqrt{25-9 - 4x^2 -12x}}$$

=

$$\frac{2}{\sqrt{16-4x^2-12x}}$$

simplifying it by dividing everything by 2

$$\frac{1}{\sqrt{8-2x^2-12x}}$$

however apparently i'm wrong again and could've simplified it further

$$\frac{1}{\sqrt{x^2-3x+4}}$$
http://www.wolframalpha.com/input/?i=d/dx+arcsin((2x+3)/5)

however I don't see how that's possible unless I have a 1/2 in the numerator

is there a property of the square root that i'm missing?

7. Aug 11, 2010

### Dick

I really think you need a nap. Suppose I told you I'd simplified 2/sqrt(16) to 1/sqrt(8). Would you believe me? Are you sure you don't see what's wrong? You've got some other problems in there, but let's stick with that for now.

8. Aug 11, 2010

### vorcil

I wouldn't believe you because 1/sqrt(8) != 2/4

before I tried to simplify
was the result

$$\frac{2}{\sqrt{16-4x^2-12x}}$$

not correct?
I thought I did that right. where would I go from here?

9. Aug 11, 2010

### Dick

That's correct. I would pull a factor of 4 out of the expression in the radical, then sqrt(4)=2 that cancels the 2 in the denominator. But this is getting silly. sqrt(a*b)=sqrt(a)*sqrt(b), right? Not sqrt(a*b)=a*sqrt(b). You MUST know that. You don't just pull factors out through powers.