You made a small mistake in the derivative of ln(x^{2}+1)^{1/2} but other than that it is correct
y = (x^{3}√(x^{2 }+ 1))/(x - 1)
What I did:
I used natural log on both sides.
I brought the exponents into the coefficient.
I also brought the (x - 1) up by doing it to the negative one exponent
Then I used the log properties.
So tell me if this is the correct way:
ln y = 3lnx + (1/2)ln(x^{2 }+ 1) - ln(x - 1)
Then I find the derivative
1/y(dy/dx) = 3/x + (1/2)(1/(x^{2 }+ 1)) - (1/(x - 1))
After that I multiplied both sides by y, then plugged in y, which is the original equation.
So would this be right?
Thanks
Hmm, so first I would put the exponent into the coefficient right?
(1/2)ln(x^{2}+1)
Um, would I then have the log go into the x and 1?
Yeah sorry, doing the derivative of 1 variable for ln is easy for me.
So if it was that (1/2)ln(x^{2}+1)
I wouldn't just do what's in the parenthesis and put it all down under 1?
Then multiply it by 1/2?
So like 1/(2(x^{2}+1))?