Derivative of the Natural Logarithm

In this page, we will calculate the derivative of the natural logarithm by using two equivalent forms to express the increment ratio: for $h\to 0$ and for $x\to x_0$. Formally, we have:

$$\underset{h\to 0}{lim}\frac{\ln (x+h)-\ln (x)}{h},\underset{x\to x_0}{lim}\frac{\ln (x)-\ln (x_0)}{x-x_0}$$

• Difference Quotient Limit as $h\to 0$
• Difference Quotient Limit as $x\to x_0$

By applying this definition to the function $\ln (x)$, we obtain:

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\ln (x+h)-\ln (x)}{h}(\ast )$$

Using the logarithmic property, we can rewrite the numerator in $(\ast )$ as:

$$\ln (x+h)-\ln (x)=\ln \left(\frac{x+h}{x}\right)$$

Thus,

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\ln (1+\frac{h}{x})}{h}$$

To simplify further, observe that this expression hides a notable limit. If we set $t=\frac{h}{x}$, then $h=xt$. Consequently, as $h\to 0$, we also have $t\to 0$. Therefore,

$$f^{\prime }(x)=\frac{1}{x}\cdot \underset{t\to 0}{lim}\frac{\ln (1+t)}{t}\stackrel{\text{Notable Limit}}{=}\frac{1}{x}\cdot 1=\frac{1}{x}$$

Thus, the derivative of $\ln (x)$ is:

$$f^{\prime }(x)=\frac{1}{x},\mathrm{\forall }x>0$$

Similarly, we calculate the limit as $x\to x_0$. Using this definition, the limit of the increment ratio is:

$$f^{\prime }(x_0)=\underset{x\to x_0}{lim}\frac{\ln (x)-\ln (x_0)}{x-x_0}$$

We apply the logarithmic property $\ln (x)-\ln (x_0)=\ln \left(\frac{x}{x_0}\right)$. The numerator becomes:

$$\ln (x)-\ln (x_0)=\ln \left(\frac{x}{x_0}\right)$$

Thus,

$$f^{\prime }(x_0)=\underset{x\to x_0}{lim}\frac{\ln \left(\frac{x}{x_0}\right)}{x-x_0}(\ast )$$

To simplify, let $u=x-x_0$, which implies that $x=x_0+u$. As $x\to x_0$, we also have $u\to 0$.

Substituting $x=x_0+u$ into the limit $(\ast )$, we get:

$$f^{\prime }(x_0)=\underset{u\to 0}{lim}\frac{\ln \left(\frac{x_0+u}{x_0}\right)}{u}$$

The argument of the logarithm can be rewritten to more easily identify the notable limit, which will allow us to compute the derivative we are seeking:

$$f^{\prime }(x_0)=\underset{u\to 0}{lim}\frac{\ln (1+\frac{u}{x_0})}{u}$$

If we set $t=\frac{u}{x_0}$, then $u=x_{0}t$. Moreover, $u\to 0$ implies that $t\to 0$:

$$f^{\prime }(x_0)=\frac{1}{x_0}\cdot \underset{t\to 0}{lim}\frac{\ln (1+t)}{t}\stackrel{\text{Notable Limit}}{=}\frac{1}{x_0}\cdot 1=\frac{1}{x_0}$$

We conclude that, as in the previous case, the derivative of $\ln (x)$ is:

$$f^{\prime }(x)=\frac{1}{x},\mathrm{\forall }x>0$$