Deriving the Derivative of the Natural Logarithm

Introduction

The natural logarithm and the natural exponential undo one another. That relationship does more than help solve equations: it lets us obtain the derivative of the logarithm from the derivative of the exponential.

The key step is the inverse-function derivative rule. In this tutorial, we state that rule, explain why it works, and apply it to derive the derivative of \(\ln x\) for positive \(x\). The restriction to positive inputs is essential: the natural logarithm is defined on \((0,\infty)\), not at zero or at negative real numbers.

The Inverse Relationship and Its Domain

The natural exponential function \(f(x)=e^x\) is strictly increasing and has range \((0,\infty)\). Consequently, it has an inverse function whose domain is \((0,\infty)\). That inverse is the natural logarithm.

These identities specify which values can be used in the inverse relationship. The exponential can take any real input, but its output is always positive. Thus \(\ln x\) is defined precisely when \(x>0\).

From the earlier result on the derivative of the natural exponential, we use \(\frac{d}{dx}e^x=e^x\). In particular, this derivative is never zero. The inverse-function derivative rule turns that fact into a formula for the derivative of its inverse.

The Inverse-Function Derivative Rule

Suppose \(f\) is differentiable near a point \(a\), is one-to-one there, and has an inverse \(f^{-1}\) on the corresponding range. If the inverse is continuous at \(f(a)\) and \(f'(a)\ne 0\), then the inverse is differentiable at \(f(a)\), with derivative equal to the reciprocal of \(f'(a)\).

The formula compares derivatives at corresponding inputs. If \(f(a)=y\), then the derivative of the inverse is evaluated at \(y\), while the derivative of \(f\) in the denominator is evaluated at \(a=f^{-1}(y)\). Keeping these two inputs distinct prevents a common substitution error.

Why the Rule Works

Here is the difference-quotient argument. Let \(y\ne f(a)\) approach \(f(a)\), and set \(x=f^{-1}(y)\). Since \(f^{-1}\) is continuous at \(f(a)\), \(x\) approaches \(a\). Also \(y=f(x)\), so the difference quotient for the inverse can be rewritten as a reciprocal:

As \(y\) approaches \(f(a)\), \(x\) approaches \(a\). By the definition of the derivative, the denominator on the right approaches \(f'(a)\). Because \(f'(a)\ne0\), taking reciprocals is valid, and the quotient approaches \(1/f'(a)\). This is exactly the derivative of \(f^{-1}\) at \(f(a)\), proving the rule.

The nonzero-derivative condition matters: if the denominator limit were zero, this reciprocal argument would not establish a finite derivative for the inverse. The assumptions of the rule are therefore part of the result, not optional details.

Deriving the Derivative of \(\ln x\)

Apply the rule to \(f(x)=e^x\). This function is differentiable, strictly increasing, and therefore has an inverse on its range. Its inverse is continuous, and \(f'(a)=e^a\), which is positive and hence nonzero for every real \(a\).

Choose any positive input \(x\) for the logarithm. Its corresponding input to the exponential is \(a=\ln x\), because \(e^{\ln x}=x\). The inverse-function rule now gives:

The final equality uses \(e^{\ln x}=x\), which holds on the logarithm’s domain. The formula is not being asserted at \(x=0\) or for \(x<0\): there is no real value \(\ln x\) at those inputs.

Worked Examples

A Common Mistake: Using the Wrong Input

The inverse rule evaluates the original function’s derivative at the input that maps to the logarithm’s input. It does not use the same number as the input to both functions.

For example, to find the derivative of \(\ln 3\) as a function at input \(x=3\), an incorrect calculation might use the exponential derivative at \(3\) and claim the answer is \(1/e^3\). That uses the wrong input: \(e^3\) is not \(3\). The correct corresponding exponential input is \(\ln 3\), because \(e^{\ln 3}=3\). Therefore:

Check: the exponential derivative at \(\ln 3\) is \(e^{\ln 3}=3\), and the inverse derivative is its reciprocal, \(1/3\). The general formula \(1/x\) records this input matching directly.

Why the Result Matters

The derivative formula describes how the natural logarithm changes as its positive input changes. Its value \(1/x\) is positive throughout the domain, so the logarithm increases there. The formula also shows that the derivative depends on the input: it is larger near zero and smaller for large positive inputs.

Most importantly for this derivation, the formula is not an independent fact to memorize without context. It follows from three ingredients: the logarithm is the inverse of the natural exponential; the exponential has derivative \(e^x\); and the derivative of an inverse is the reciprocal derivative at the corresponding input. The correspondence \(e^{\ln x}=x\) completes the argument.