What is the derivative of #ln(1/x)#?

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Answer 1 · 2016-09-01T14:25:05

The derivative is #-1/x#

Here are two ways to find the derivative:

Method 1 relies on knowing about exponents and logarithms.

We'll use #1/x = x^-1# and #ln(a^b) = blna# and #d/dx(cf(x)) = c d/dx(f(x))#

We get:

#d/dx(ln(1/x)) = d/dx(ln(x^-1))#

# = d/dx(-1lnx) = -1 d/dx(lnx)#

# = -1/x#.

Method 2 uses the chain rule.

#d/dx(lnu) = 1/u * (du)/dx# and

#d/dx(1/x) = d/dx(x^-1) = -x^-2 = -1/x^2#.

We get

#d/dx(ln(1/x)) = 1/(1/x) * d/dx(1/x)#

# = x * (-1/x^2)#

# = -1/x#