Hi please help deriving this question:see the image for the equation. deriv.bmp
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
If y is the log of a function of x
$\displaystyle y=ln(g(x))$
Substitute for the function in the log and use the chain rule.
Let u= g(x), then
$\displaystyle y=ln(u)$
$\displaystyle \frac{du}{dx}=g'(x)$
$\displaystyle \frac{dy}{du}=\frac{1}{u}=\frac{1}{g(x)}$
$\displaystyle \frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=\frac{1}{g(x)} g'(x)$
So for your problem $\displaystyle f'(x)=\frac{-3x^2-5}{-5-5x-x^3}$