Simplification problems (I hope)

[Tjvelcro]

Homework Statement

1. Differentiate g(x)= ln(x(x^2 - 1)^1/2

2. Differentiate g(x)= x^2(ln(2x))

Homework Equations

Law of logs and product rule

The Attempt at a Solution

1. I simplified to g'(x) = 1/x + (1/2)(1/(X^2 - 1)(2x) then
g'(x) = 1/x + 2x/(2(x^2 - 1))
Answer however is g'(x)= (2x^2-1)/x(x^2-1)

2. I simplified to g'(x) = 2xln(2x) + x(1/2x)
Answer however is g'(x) = x + 2xln(2x) I am not sure how to get rid of the 2 in the denominator of the 2nd term

Tjvelcro

 

[xcvxcvvc]

g'(x) = \frac{1}{x} + \frac{2x}{2(x^2 - 1)}
g'(x) = \frac{2(x^2-1)}{2x(x^2-1)} + \frac{2x^2}{2x(x^2 - 1)}
g'(x) = \frac{2(x^2 - 1) + 2x^2}{2x(x^2-1)}
g'(x) = \frac{2x^2 - 2 + 2x^2}{2x(x^2-1)}
g'(x) = \frac{4x^2 - 2}{2x(x^2-1)}
g'(x) = \frac{2x^2 - 1}{x(x^2-1)}

g'(x) = 2xln(2x) + x(1/2x)

That is wrong. Did you mistakenly type the x in the denominator of that second term? With or without that extra x, the answer is still incorrect. Post what you did.

 

[Tjvelcro]

So for the second problem

Differentiate g(x)= x^2(ln(2x))
I used product rule.

g'(x) = 2x(ln(2x)) + (x^2)(2/2x)
g'(x) = 2x(ln(2x)) + (x) This is where I made my mistake, I get confused with y'lnx=1/x and y'[lng(x)] = g'(x)/g(x)

Thanks

Tjvelcro