## Derivatives of Logarithmic Functions

Implicit differentiation, which we explored in the last section, can also be employed to find the derivatives of logarithmic functions, which are of the form $y = \log_a{x}$. This also includes the natural logarithmic function $y = \ln{x}$.

## Proving $\frac{d}{dx} (\log_a{x}) = \frac{1}{x \ln{a}}$¶

Taking advantage of the fact that $y = \log_a{x}$ can be rewritten as an exponential equation, $a^y = x$, we can state the derivative of $\log_a{x}$ as:

$$\frac{d}{dx} (\log_a{x}) = \frac{1}{x \ln{a}}$$

This can be proved by applying implicit differentiation. First we find the deriative of $y = a^x$. Start by taking the $\ln{}$ of both sides of the equation:

$$\ln{y} = \ln{a^x}$$

Then exponentiate both sides:

$$e^{\ln{y}} = e^{\ln{a^x}}$$

As $a^{\ln{x}}$ = $x \ln{e}$, and $\ln{e} = 1$, we can simplify the left side of the equation to remove the exponent and natural log.

$$y = e^{\ln{a^x}}$$

Proceed to find the derivative by using the chain rule:

$$\frac{dx}{dy} = e^{\ln{a^x}}, \qquad y = e^u, \qquad u = \ln{a^x}$$

$$\frac{dy}{du} = e^u, \qquad \frac{du}{dx} = \ln{a}$$

$$\ln{a} e^u = \ln{a} e^{\ln{a}}$$

Just as $\ln{e} = 1$, $e^{\ln{x}} = x$, thus:

$$\frac{dy}{dx} \ln{a} \space a^x$$

Returning to the derivative of $y = \log_a{x}$, rewrite the equation:

$$a^y = x$$

Now that we know $\frac{dy}{dx} \ln{a} \space a^x$, we can derive the equation implicitly with respect to $x$:

$$\frac{d}{dx} a^y = \frac{d}{dx} x$$

$$a^y \ln{a} \frac{dy}{dx} = 1$$

Then, dividing by $a^y \ln{a}$:

$$\frac{d}{dx} \log_a{x} = \frac{1}{x \ln{a}}$$

## Proving $\frac{d}{dx} (\ln{x}) = \frac{1}{x}$¶

Start with $y = \ln{x}$. Exponentate both sides of the equation:

$$e^y = e^{\ln{x}}$$

Recalling that $e^{\ln{x}} = x$:

$$e^y = x$$

Applying implicit differentiation:

$$\frac{dy}{dx} e^y = \frac{d}{dx} x$$

$$e^y \frac{dy}{dx} = 1$$

Using the fact that $e^y = x$, we can substitute $x$ for $e^y$:

$$x \frac{dy}{dx} = 1$$

Divide by $x$:

$$\frac{dy}{dx} = \frac{1}{x}$$

## Examples¶

In [26]:
from sympy import symbols, diff, simplify, sin, cos, log, tan, sqrt, init_printing
from mpmath import ln, e

init_printing()
x = symbols('x')
y = symbols('y')


### Example 1: Calculate the derivative of $f(x) = \sin{\ln{x}}$¶

Apply the Chain Rule:

$$y = \sin{u}, \qquad u = \ln{x}$$

As we proved above, $\frac{du}{dx} = \frac{1}{x}$, therefore:

$$\frac{dy}{dx} = \frac{1}{x} \cos{\ln{x}}$$

In [ ]:
diff(sin(ln(x)))


## Example 2: Find the derivative of the function $f(x) = \log_2{(1 - 3x)}$¶

Using the fact that $\frac{d}{dx} (\log_a{x}) = \frac{1}{x \ln{a}}$, apply the Chain Rule:

$$y = \log_2{u}, \qquad u = 1 - 3x$$

$$\frac{dy}{du} = \frac{1}{u \ln{2}}, \qquad \frac{du}{dx} = 3$$

$$\frac{dy}{dx} = \frac{3}{(1 - 3x) \ln{2}}$$

In [9]:
diff(log((1 - 3 * x), 2))

Out[9]:
$$- \frac{3}{\left(- 3 x + 1\right) \log{\left (2 \right )}}$$

### Example 3: Find the derivative to the function $\sqrt[5]{\ln{x}}$¶

First, rewrite the equation as $\sqrt[5]{\ln{x}} = (\ln{x})^{\frac{1}{5}}$. Then, apply the Chain Rule:

$$y = u^\frac{1}{5}, \qquad u = \ln{x}$$

$$\frac{dy}{du} = \frac{1}{5} (\ln{x})^{-\frac{4}{5}}, \qquad \frac{du}{dx} = \frac{1}{x}$$

$$\frac{dy}{dx} = \frac{1}{x} \space \frac{1}{5} (\ln{x})^{-\frac{4}{5}} = \frac{(\ln{x})^{-\frac{4}{5}}}{5x} = \frac{1}{5x \ln^\frac{4}{5}{x}}$$

In [13]:
diff((log(x)) ** (1/5))

Out[13]:
$$\frac{0.2}{x \log^{0.8}{\left (x \right )}}$$

### Example 4: Find the derivative of $f(x) = \sin{x} \ln{5x}$¶

Start by using the Product rule, $u \frac{dv}{dx} + v \frac{du}{dx}$:

$$u = \sin{x}, \qquad v = \ln{5x}$$

$$\frac{dy}{dx} = (\sin{x}) \frac{dv}{dx} \ln{5x} + (\ln{5x}) \frac{du}{dx} \sin{x}$$

$$\frac{dy}{dx} = \frac{\sin{x}}{x} + \ln{5x} \cos{x}$$

In [28]:
diff(sin(x) * log(5 * x))

Out[28]:
$$\log{\left (5 x \right )} \cos{\left (x \right )} + \frac{1}{x} \sin{\left (x \right )}$$

### Example 5: Find the derivative of the function $F(x) = \ln{\frac{(2x + 1)^3}{(3x - 1)^4}}$¶

Here, rather than applying the Chain Rule and then the Quotient Rule to the inner equation, which would result in a very lengthy and tedious derivation, we can take advantage of one of the logarithmic identities, $\ln{ab} = \ln{a} + \ln{b}$. Thus, we can rewrite the function as:

$$F(x) = \ln{(2x + 1)^3} + \ln{\frac{1}{(3x - 1)^4}}$$

With the rewritten function, we can leverage another logarithmic identity, $\ln{a^b} = b \ln{a}$ to simplify the derivation. We can therefore rewrite the function again as:

$$F(x) = 3 \ln{(2x + 1)} - 4 \ln{(3x - 1)}$$

Then, proceed to differentiate the function:

$$\frac{dy}{dx} = 3 \frac{d}{dx} \ln{(2x + 1)} - 4 \frac{d}{dx} \ln{(3x - 1)}$$

$$\frac{dy}{dx} = 3 \frac{2}{2x + 1} - 4 \frac{3}{3x - 1} = \frac{6}{2x + 1} - \frac{12}{3x - 1}$$

In [35]:
simplify(diff(log((2 * x + 1) ** 3 / (3 * x - 1) ** 4)))

Out[35]:
$$- \frac{6 x + 18}{6 x^{2} + x - 1}$$

## References¶

Stewart, J. (2007). Essential calculus: Early transcendentals. Belmont, CA: Thomson Higher Education.