# Math Help - Logarithmic Differentiation: Finding derivative

1. ## Logarithmic Differentiation: Finding derivative

y = (x3√(x2 + 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(x2 + 1) - ln(x - 1)

Then I find the derivative
1/y(dy/dx) = 3/x + (1/2)(1/(x2 + 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

2. ## Re: Logarithmic Differentiation: Finding derivative

You made a small mistake in the derivative of ln(x2+1)1/2 but other than that it is correct

3. ## Re: Logarithmic Differentiation: Finding derivative

Originally Posted by Shakarri
You made a small mistake in the derivative of ln(x2+1)1/2 but other than that it is correct
Hmm, so first I would put the exponent into the coefficient right?
(1/2)ln(x2+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(x2+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(x2​+1))?

4. ## Re: Logarithmic Differentiation: Finding derivative

$v= ln(x^2+1)$

let $u=(x^2+1)$

$\frac{du}{dx}=2x$

$v=ln(u)$

$\frac{dv}{du}=\frac{1}{u}$

$\frac{dv}{dx}= \frac{du}{dx}\cdot \frac{dv}{du}$

$\frac{dv}{dx}=2x \cdot \frac{1}{u}= \frac{2x}{(x^2+1)}$

5. ## Re: Logarithmic Differentiation: Finding derivative

Originally Posted by Shakarri
$v= ln(x^2+1)$

let $u=(x^2+1)$

$\frac{du}{dx}=2x$

$v=ln(u)$

$\frac{dv}{du}=\frac{1}{u}$

$\frac{dv}{dx}= \frac{du}{dx}\cdot \frac{dv}{du}$

$\frac{dv}{dx}=2x \cdot \frac{1}{u}= \frac{2x}{(x^2+1)}$
Oh right, now I remember!
So it's the derivative of that over the original!

Thanks!

Then after that you would multiply it by 1/2 right?
Which will be 2x/(2(x2​+1))