# Thread: Use logarithmic differentiation to find the derivative of y= (x√(x+1))/(2 + x^4)

1. ## Use logarithmic differentiation to find the derivative of y= (x√(x+1))/(2 + x^4)

$\displaystyle y= \frac{x\sqrt{x+1}}{(2 + x^4)}$
$\displaystyle lny= ln \frac{x\sqrt{x+1}}{(2 + x^4)}$
$\displaystyle lny= ln({x\sqrt{x+1}) - ln{(2 + x^4)}$

Is that correct so far? And any advice for what to do after...I've never done a question like this and it's confusing me!

2. Write it as $\displaystyle \ln (y) = \ln (x) + \frac{1}{2}\ln (x + 1) - \ln (2 + x^4 )$

3. Thanks..I'm trying to work it out.

$\displaystyle \frac{y'}{y}= \frac{1}{x} + \frac{1}{2x+2}-\frac{1}{2+x^4}$

Am I on the right track or way off base?

4. You missed a $\displaystyle 4x^3$ is the last term.
Now multiply all by y.

5. I don't understand your last term. Given that $\displaystyle \dfrac{d}{dx} \ln[f(x)] = \dfrac{f'(x)}{f(x)}$

In your case $\displaystyle \dfrac{d}{dx} \ln(2+x^2) = \dfrac{4x^3}{2+x^4}$

Rest of it is fine though, remember that you made a y sub when multiplying by y.

6. Thanks so much for the help
I think I have the solution figured out now.
$\displaystyle y'= -\frac{(5x^5 + 6x^4 - 6x - 4)y}{2x(x + 1)(x^4 + 2)}$

7. Quite frankly, I prefer the answer form $\displaystyle \displaystyle y' = \frac{{x\sqrt {x + 1} }}{{2 + x^4 }}\left\{ {\frac{1}{x} + \frac{1} {{2\left( {x + 1} \right)}} - \frac{{4x^3 }} {{2 + x^4 }}} \right\}$