Question

Natural logarithm - find f'(x) if f(x) is the given expression.

f(x) = 1/ln x + ln 1/x

Answer 1

I'm just looking to find out how to break down ln x when using logarithmic differentiation.

Answer 2

for 1/ln x, use the chain rule d/dy(f(g(x))) = f'(g(x))*g'(x)
(ln x)^-1.
f = x^-1, g = ln x
f'= -1/x^2, g' = 1/x
f'(g)*g' = -1/(ln x)^2 * 1/x = -1/ ( x Ln(x)^2)

Answer 3

for ln(1/x) use properties of logs to break it up
ln(1/x) = ln(1) - ln(x) = -ln(x)
The derivative is -1/x