Differentiation of a logarithm

Let^5.3 $y = log_av$.

Differentiating by the General Rule (§4.7), considering $v$ as the independent variable, we have

• FIRST STEP. $\ y + \Delta y = \log_a(v + \Delta v)$ .
• SECOND STEP^5.4.
  $$\begin{array}{ll} \Delta y &= \log_a(v + \Delta v) - \log_a... ...& = \log_a \left ( 1 + \frac{\Delta v}{v} \right ). \end{array}$$
  by item (8), §1.1.

• THIRD STEP.
  $$\begin{array}{ll} \frac{\Delta y}{\Delta x} &= \frac{1}{\Del... ...+ \frac{\Delta v}{v} \right )^{\frac{v}{\Delta v}}. \end{array}$$
  [Dividing the logarithm by $v$ and at the same time multiplying the exponent of the parenthesis by $v$ changes the form of the expression but not its value (see item (9), §1.1.]

• FOURTH STEP. $\frac{dy}{dv} = \frac{1}{v} \log_a e$. [When $\Delta v \to 0$ $\frac{\Delta v}{v} \to 0$. Therefore $\lim_{\Delta v \to 0} \left ( 1 + \frac{\Delta v}{v} \right )^{\frac{v}{\Delta v}} = e$, from §3.11, placing $x = \frac{\Delta v}{v}$.]

Hence

$$\frac{dy}{dv} = \frac{d}{dv} \left ( \log_a v \right ) = \log_a e \cdot \frac{1}{v}.$$ (5.3)

Since $v$ is a function of $x$ and it is required to differentiate $\log_a v$ with respect to $x$, we must use formula (5.1), for differentiating a function of a function, namely,

$$\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}.$$

Substituting the value of $\frac{dy}{dv}$ from (5.3), we get

$$\frac{dy}{dx} = \log_a e \cdot \frac{1}{v} \cdot \frac{dv}{dx}.$$

Therefore, $\frac{d}{dx} (\log_a x) = \log_s e \cdot \frac{\frac{dv}{dx}}{v}$ (equation (VIII) above). When $a = e$, $\log_a e = log_e e = 1$, and (VIII) becomes $\frac{d}{dx} (\log v) = \frac{\frac{dv}{dx}}{v}$ (equation (VIIIa) above).

The derivative of the logarithm of a function is equal to the product of the modulus^5.5of the system of logarithms and the derivative of the function, divided by the function.