# Thread: derivative involving natural log

1. ## derivative involving natural log

so the function is y= ln(arcsec(3x+1))

i know that the derivative of lnx is 1/x

so, based on this fact, and the chain rule, i said:

y'= 1/arcsec(3x+1) * 1/x sqrt(x^2-1) * 3

i know that's not the neatest way to write it.. can someone tell me if this is correct?

2. Originally Posted by buttonbear
so the function is y= ln(arcsec(3x+1))

i know that the derivative of lnx is 1/x

so, based on this fact, and the chain rule, i said:

y'= 1/arcsec(3x+1) * 1/x sqrt(x^2-1) * 3

i know that's not the neatest way to write it.. can someone tell me if this is correct?
$\displaystyle y = \ln{[arcsec{(3x + 1)}]}$

Let $\displaystyle u = arcsec{(3x + 1)}$ so $\displaystyle y = \ln{u}$

$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$

$\displaystyle \frac{dy}{dx} = u^{-1}\frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = [arcsec{(3x+1)}]^{-1}\frac{du}{dx}$.

Now let's find $\displaystyle \frac{du}{dx}$.

Let $\displaystyle v = 3x + 1$ so $\displaystyle u = arcsec{\,v}$.

$\displaystyle \frac{du}{dx} = \frac{dv}{dx}\,\frac{du}{dv}$

$\displaystyle \frac{du}{dx} = 3\frac{1}{v^2\sqrt{1 - \frac{1}{v^2}}}$

$\displaystyle \frac{du}{dx} = \frac{3}{(2x + 1)^2\sqrt{1 - \frac{1}{(3x + 1)^2}}}$

So $\displaystyle \frac{dy}{dx} = \frac{3}{[arcsec{(3x+1)}](2x + 1)^2\sqrt{1 - \frac{1}{(3x + 1)^2}}}$.

3. now i'm really confused

edit:

okay, now that i can see the symbols it makes more sense
i'm not really sure why my method didn't work, though..

can you just explain where the (2x+1) came from?