# Thread: TEST derivative using y= lnx

2. ## Re: TEST derivative using y= lnx

i do know that one of the rules involving natural logarithms is: y=lnx which derived is y'= 1/x . But I'mstill not sure. This is my teacher's solution:y'= -5-3x^2 / (-5-5x-x^3)

3. ## Re: TEST derivative using y= lnx

y =ln(-5-5x-x^3)

derivative of ln(x) is y' = 1/(x)*(x)'

y'= (ln(-5-5x-x^3))'*(-5-5x-x^3)' so, you have 1/(-5-5x-x^3)*(5x-3x^2) =(5x-3x^2)/(-5-5x-x^3)

dokrbb

4. ## Re: TEST derivative using y= lnx

If y is the log of a function of x

$y=ln(g(x))$

Substitute for the function in the log and use the chain rule.
Let u= g(x), then
$y=ln(u)$

$\frac{du}{dx}=g'(x)$

$\frac{dy}{du}=\frac{1}{u}=\frac{1}{g(x)}$

$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=\frac{1}{g(x)} g'(x)$

So for your problem $f'(x)=\frac{-3x^2-5}{-5-5x-x^3}$

thank you