Why Do I Get a Different Result When Differentiating ln(x + sqrt(x^2-1))?

UsernameValid
Messages
12
Reaction score
0
How do I find the derivative of (x)=ln(x+(x^2-1)1/2)

The answer is suppose to be 1/(x2-1). But I keep ending up with 2x/(x2-1).
 
Last edited:
Physics news on Phys.org
UsernameValid said:
How do I find the derivative of (x)=ln(x+(x^2-1)1/2)
Is this supposed to be f(x) = ...
UsernameValid said:
The answer is suppose to be 1/(x2-1). But I keep ending up with 2x/(x2-1).

Show us what you did.

Also, do not delete the three parts of the homework template. They are there for a reason.
 
Well, the problem says h(x).


(d/dx)ln(x+(x2-1)1/2*(d/dx)(x+(x2-1)1/2*(d/dx)(x2-1)1/2*(d/dx)(x2-1).

1/(x+(x2-1)1/2 * 1+[x/(x2-1)]1/2 * x/(x2-1)1/2 * 2x

Which actually gets me 2x2/(x2-1)
 
UsernameValid said:
(d/dx)ln(x+(x2-1)1/2*(d/dx)(x+(x2-1)1/2*(d/dx)(x2-1)1/2*(d/dx)(x2-1).
I can't decipher that line. Left and right parentheses don't match up. Is there an equals sign missing? If not, I don't understand why there are any log terms in there.
 
UsernameValid said:
Well, the problem says h(x).
How was I to tell? You wrote this
How do I find the derivative of (x)=ln(x+(x^2-1)1/2)
UsernameValid said:
(d/dx)ln(x+(x2-1)1/2*(d/dx)(x+(x2-1)1/2*(d/dx)(x2-1)1/2*(d/dx)(x2-1).
There are actually too many "d/dx" operators in there, although I get what you're trying to do. Your task is to do this differentiation:

d/dx(ln(x+(x2-1)1/2)

$$= \frac{1}{x + (x^2 - 1)^{1/2}} \cdot d/dx(x + (x^2 - 1)^{1/2})$$
$$= \frac{1}{x + (x^2 - 1)^{1/2}} \cdot (1 + d/dx[(x^2 - 1)^{1/2}]$$
and so on, whittling away at it a little at a time.
UsernameValid said:
1/(x+(x2-1)1/2 * 1+[x/(x2-1)]1/2 * x/(x2-1)1/2 * 2x

Which actually gets me 2x2/(x2-1)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top