Differentiation of logarithmic and exponential functions

a) Derivatives of natural logarithmic functions

Given $y = \ln x$, in exponential form, it's $x = e^y$.

Differentiating with respect to $x$:

$\frac{d(e^y)}{dx} = 1$

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

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

Since $x = e^y$:

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

Generally, differentiating natural logarithms of the form:

a) $y = \ln(ax)$ b) $y = \ln(ax + b)$

Use the chain rule:

a) $y = \ln(ax)$ ($ax > 0$)

$\text{Let } u = ax$

$\frac{du}{dx} = a$

$y = \ln u$

$\frac{dy}{du} = \frac{1}{u}$

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \frac{1}{u} \cdot a = \frac{a}{u}$

Since $u = ax$:

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

b) $y = \ln(ax + b)$ ($ax + b > 0$)

$\text{Let } u = ax + b$

$\frac{du}{dx} = a$

$y = \ln u$

$\frac{dy}{du} = \frac{1}{u}$

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \frac{1}{u} \cdot a = \frac{a}{u}$

Since $u = ax + b$:

$\frac{dy}{dx} = \frac{a}{ax + b}$

Generally:

$\frac{d}{dx}\bigl[\ln f(x)\bigr] = \frac{f'(x)}{f(x)}$

Example 9.10: Finding first derivatives

a) $f(x) = \ln(3x + 1)$ b) $g(x) = \ln(2x - 3)$

Solution:

a) Let $u = 3x + 1$

$\frac{du}{dx} = 3$

$\frac{dy}{du} = \frac{1}{u}$

$\frac{dy}{dx} = \frac{1}{u} \cdot 3 = \frac{3}{3x + 1}$

b) Let $u = 2x - 3$

$\frac{du}{dx} = 2$

$\frac{dy}{du} = \frac{1}{u}$

$\frac{dy}{dx} = \frac{1}{u} \cdot 2 = \frac{2}{2x - 3}$

b) Derivatives of common logarithmic functions

Steps:

1. Convert the common logarithm to exponential form.
2. Convert the exponential form to natural logarithm form.
3. Differentiate the natural logarithm.

Consider $y = \log_a x$. In exponential form: $x = a^y$.

Converting to natural logarithms:

$\ln x = \ln a^y$

$\ln x = y \ln a$

Differentiating implicitly with respect to $x$:

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

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

Alternatively, by change of base to $e$:

$\log_a x = \frac{\ln x}{\ln a}$

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

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

Example 9.11: Differentiating common logarithms

a) $f(x) = \log_{10} x$ b) $g(x) = \log_{10}(x + e^{2x+3})$ c) $h(x) = \log_{10}(x^2 + 4x + 5)$

Solution:

a) $y = \log_{10} x$

$\ln y = \frac{\ln x}{\ln 10}$

$\frac{dy}{dx} = \frac{1}{x \ln 10}$

b) $y = \log_{10}(x + e^{2x+3})$

$\ln y = \frac{\ln(x + e^{2x+3})}{\ln 10}$

$\frac{dy}{dx} = \frac{1}{\ln 10} \cdot \frac{1 + 2e^{2x+3}}{x + e^{2x+3}}$

c) $y = \log_{10}(x^2 + 4x + 5)$

$\ln y = \frac{\ln(x^2 + 4x + 5)}{\ln 10}$

$\frac{dy}{dx} = \frac{1}{\ln 10} \cdot \frac{2x + 4}{x^2 + 4x + 5}$

c) Derivatives of exponential functions ($a^x$)

Differentiating $f(x) = a^x$ ($a \neq 0$) using natural logarithms:

$y = a^x$

$\ln y = \ln a^x$

$\ln y = x \ln a$

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

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

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

If $a = e$:

$\frac{dy}{dx} = e^x \ln e$

Since $\ln e = 1$:

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

Example 9.12: Finding dy/dx

$y = 2^x$

Solution:

$\ln y = x \ln 2$

$\frac{1}{y} \frac{dy}{dx} = \ln 2$

$\frac{dy}{dx} = y \ln 2$

$\frac{dy}{dx} = 2^x \ln 2$

Example 9.13: Finding the derivative

$f(x) = 10^{x^2}$

Solution:

$y = 10^{x^2}$

$\ln y = x^2 \ln 10$

$\frac{1}{y} \frac{dy}{dx} = 2x \ln 10$

$\frac{dy}{dx} = y \cdot 2x \ln 10$

$\frac{dy}{dx} = 10^{x^2} \cdot 2x \ln 10$

$\frac{dy}{dx} = 2x \cdot 10^{x^2} \ln 10$