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

2 Answers
Aug 22, 2014

You can find this derivative by applying the Chain Rule, with #cosx# as the inner function, and #lnx# as the outer function.

Process:

To apply the chain rule, we first find the derivative of the outer function, #lnu#, with #u = cosx#. Remember that the derivative of #lnu = 1/u = 1/cosx#.

Now we just need to find the derivative of the inner function, #cosx#, and multiply it by the derivative of the outer function we just found.

Since the derivative of #cosx# is (#-sinx#), we end up with:

#dy/dx = (1/cosx) * (-sinx) = (-sinx/cosx) = -tanx#.

A shorter way to do these is to just know that the derivative of a #ln(u)#-type function is the derivative of the inside over the original of what's inside.

Jun 4, 2017

#dy/dx=-tanx#

Explanation:

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

#• d/dx(ln(f(x)))=(f'(x))/(f(x))#

#rArrdy/dx=(-sinx)/(cosx)=-tanx#