Second derivative of chain rule in chain rule in achain rule

[rbcobra]

I cant seem to get this one:

Take to the second derivative:
(1+ln cosx)^(-1)

first derivative (i think):
[-1(1+ln cosx)^(-2)] *(1+(cosx)^(-1)) * (-sinx) so therefore when x=0, the first derivative equals 0

second derivative (i know it is supposed to equal 1, however I cannot seem to get there)

[2(1+ln cosx)^(-3)] * #now do I break the inside apart again? (that gives me zero when x=0, because sin0=0, or continue from the first derivative?)# * (-cosx)^(-2) * (-cosx)

Any help would be much appreciated!
Cheers,
X

 

[galactus]

The first derivative isn't that bad by using the chain rule.

\(\displaystyle (1+ln(cos(x)))^{-1}\)

Derivative of outside is merely \(\displaystyle -(1+ln(cos(x)))^{-2}\)

The derivative of the inside is just the derivative of ln(cos(x)) because the derivative of 1 is 0. Which is \(\displaystyle \frac{1}{cos(x)}\cdot -sin(x)=-tan(x)\)

So, we have \(\displaystyle \frac{tan(x)}{(1+ln(cos(x)))^{2}}\)

Now, we can use the product rule or quotient rule. Try \(\displaystyle \underbrace{tan(x)}_{\text{f(x)}}\cdot \underbrace{(1+ln(cos(x)))^{-2}}_{\text{g(x)}}\)

\(\displaystyle \underbrace{tan(x)}_{\text{f(x)}}\cdot \overbrace{\frac{2tan(x)}{(1+ln(cos(x)))^{3}}}^{\text{g'(x)}}+\underbrace{(1+ln(cos(x)))^{-2}}_{\text{g(x)}}\cdot \overbrace{sec^{2}(x)}^{\text{f'(x)}}\)

Tidy up.