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

2 Answers
May 19, 2018

Answer:

#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#

May 19, 2018

Answer:

#"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#