Question

How do you differentiate `y =sin(ln(cos x))` using the chain rule?

Answer 1

`dy/dx = - cos (ln(cosx)) . tanx`

Explanation:

Differentiate using the 'chain rule' :

`dy/dx = cos(ln(cosx)) .d/dx (ln(cosx))`

`dy/dx = cos(ln(cosx)).( 1/cosx . d/dx(cosx))`

`dy/dx = cos(ln(cosx)) . (1/cosx .(- sinx))`

`dy/dx = cos(ln(cosx)) . (- sinx/cosx )`

`rArr dy/dx = - cos(ln(cosx)). tanx`