How do you differentiate `y=ln(cos ^5(3x^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)`
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