This problem uses the chain rule over and over.

`y=ln(cos^5(3x^4))`

Write this as

`y=ln(f(x))` where

`f(x)=cos^5(3x^4)`

The chain rule says the derivative is:

`y'=(1/(f(x))) f'(x)=1/(cos^5(3x^4)) f'(x)`

Now we need the derivative of `f(x)`

But we can think of `f(x)=(g(x))^5`

where `g(x)=cos(3x^4)`

` ` Chain rule says:

`f'(x)=5(g(x))^4 g'(x) = 5(cos(3x^4))^4...

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This problem uses the chain rule over and over.

`y=ln(cos^5(3x^4))`

Write this as

`y=ln(f(x))` where

`f(x)=cos^5(3x^4)`

The chain rule says the derivative is:

`y'=(1/(f(x))) f'(x)=1/(cos^5(3x^4)) f'(x)`

Now we need the derivative of `f(x)`

But we can think of `f(x)=(g(x))^5`

where `g(x)=cos(3x^4)`

` ` Chain rule says:

`f'(x)=5(g(x))^4 g'(x) = 5(cos(3x^4))^4 g'(x)`

So

`y'=1/(cos^5(3x^4)) * 5 (cos^4(3x^4)) g'(x)=`

`5/(cos(3x^4)) g'(x)`

Now we need the derivative `g'(x)`

But we can write `g(x)` as

`g(x)=cos(h(x))`

where `h(x)=3x^4`

Chain rule says

`g'(x)=-sin(h(x)) h'(x) = -sin(3x^4) h'(x)`

So

`y'=5*1/(cos(3x^4)) * -sin(3x^4) h'(x)`

`=-5 tan(3x^4) h'(x)`

`h'(x)=12 x^3` using the power rule

So:

`y'=-5 tan(3x^4)*12x^3=-60x^3 tan(3x^4)`