Question

How do you find dy/dx given `y=ln(cos x)`?

Answer 1

`dy/dx=-tanx`

Explanation:

Given:
`y=ln(cosx)`
Let
`t=cosx`
Then,
`y=ln(t)`
By chain rule
`dy/dx=dy/dt.dt/dx`
`y=lnt`
`dy/dt=1/t`
`t=cosx`
`dt/dx=-sinx`
Thus
`dy/dx=(1/t).(-sinx)`
`t=cosx`
`dy/dx=(1/cosx).(-sinx)`
`dy/dx=-sinx/cosx`
`dy/dx=-tanx`

Answer 2

`dy/dx=-tanx`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"Given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`y=ln(cosx)`

`rArrdy/dx=1/cosx xxd/dx(cosx)`

`color(white)(rArrdy/dx)=-sinx/cosx=-tanx`