What is the derivative of #f(x) = ln(sinx))#?

2 Answers
Apr 22, 2018

#f'(x)=cotx#

Explanation:

We'll apply the Chain Rule, which, when applied to logarithms, tells us that if #u# is some function in terms of #x,# then

#d/dxlnu=1/u*(du)/dx#

Here, we see #u=sinx,# so

#f'(x)=1/sinx*d/dxsinx#

#f'(x)=cosx/sinx#

#f'(x)=cotx#

Apr 22, 2018

#f'(x)=cotx#

Explanation:

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

#"Given "f(x)=g(h(x))" then"#

#f'(x)=g'(h(x))xxh'(x)larrcolor(blue)"chain rule"#

#"here "f(x)=ln(sinx)#

#rArrf'(x)=1/sinx xxd/dx(sinx)=cosx/sinx=cotx#