Question

How do you find the derivative of `y=ln(sin(x))` ?

Answer 1

`cotx`

Explanation:

We use the chain rule, which states that,

`dy/dx=dy/(du)*(du)/dx`

Let `u=sinx,:.(du)/dx=cosx`.

Then, `y=lnu,dy/(du)=1/u`.

Combining, we get:

`dy/dx=1/u*cosx`

`=cosx/u`

Substituting back `u=sinx`, we get:

`=cosx/sinx`

Notice how it equals to:

`=(sinx/cosx)^-1`

But `sinx/cosx=tanx`, so we get:

`=(tanx)^-1`

`=1/tanx`

`=cotx`

Answer 2

`"dy/dx=cotx`

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(sinx)`

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

`color(white)(rArrdy/dx)=cosx/sinx=cotx`