Home > Calculus > Differentiating Logarithmic Functions with Base e

Why is the derivative of #ln -x = 1/x#?

Answer 1

Cameron Alexander

This is because the chain rule says:

#y=f(g(x))=f'(g(x))*g'(x)#.

So:

#y=ln(-x)rArry'=1/-x*(-1)=1/x#.

The #1/-x# is the derivative of the logarithmic function and the #-1# is the derivative of #-x#.

For the same reason in this integral:

#int1/xdx=ln|x|+c#

we have to put the absolue value to #x#, because we want to write all the primitive functions.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Anctil

The derivative of ln(-x) is 1/x because ln(-x) is the natural logarithm of -x, and its derivative is the reciprocal of the argument, which is 1/x. This is a consequence of the chain rule and the derivative of the natural logarithm function.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.