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

2 Answers

#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#

Mar 9, 2018

#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#