]> Exponential and Logarithmic Functions and Calculus

Exponential and Logarithmic Functions and Calculus

Exponential Functions: Differentiation

The simplest one yet: $\frac{d}{d x} (e^{x}) = e^{x}$ . The chain rule version: $\frac{d}{d x} (e^{f (x)}) = e^{f (x)} \cdot f^{'} (x)$ .

If you need to take the derivative of some other exponential function, you'll need to use logarithmic differentiation…

Logarithmic Functions: Differentiation

The Definition

Not as easy as the exponential, but darn close: $\frac{d}{d x} [ln (x)] = \frac{1}{x}$ . The chain rule version: $\frac{d}{d x} [ln (f (x))] = \frac{1}{f (x)} \cdot f^{'} (x)$ .

This is actually the definition of the natural log function—or a form of that definition, at least.

If you want to take the derivative of a logarithm with some other base, then you'll need to use the change of base formula to rewrite it with natural logs.

The Technique

Sometimes, the use of logarithms can help you to take derivatives that are otherwise mighty ugly. This is because of the fantastic property of logarithms that allows you to bring exponents down as coefficients. Also, the product/quotient properties come in handy.

The only catch is that in using logarithms, you'll technically need to use implicit differentiation, which means that you might have some y terms that need to be expanded with the original definition…

Examples:

[1.] Find $g^{'} (x)$ if $g (x) = {log}_{10} (x)$ .

$g (x) = {log}_{10} (x) = \frac{ln (x)}{ln (10)}$ ⇒ $g^{'} (x) = \frac{1}{ln (10)} \cdot \frac{1}{x} = \frac{1}{x \cdot ln (10)}$ .

[2.] Find y′ if y = 2^x.

First, take the log of both sides: $ln (y) = ln (2^{x})$ . Now pull that exponent down: $ln (y) = x \cdot ln (2)$ . Remember that ln(2) is a constant! Take the derivative, remembering that you are really doing implicit differentiation: $\frac{1}{y} \cdot y^{'} = ln (2)$ . Solve for y′: $y^{'} = y \cdot ln (2)$ . Finally, replace the y with its original definition—in this case, y = 2^x: $y^{'} = 2^{x} \cdot ln (2)$ .

[3.] Find $f^{'} (x)$ if $f (x) = \frac{x}{\sqrt{x^{2} + 1}}$ .

$y = \frac{x}{\sqrt{x^{2} + 1}}$ ⇒ $(y) = ln (\frac{x}{\sqrt{x^{2} + 1}})$ ⇒ $ln (y) = ln (x) - ln (\sqrt{x^{2} + 1})$ ⇒ $ln (y) = ln (x) - \frac{1}{2} ln (x^{2} + 1)$ . Now take the derivative! $\frac{1}{y} \cdot y^{'} = \frac{1}{x} - \frac{1}{2} (\frac{1}{x^{2} + 1}) (2 x)$ . Solve and substitute: $y^{'} = y \cdot [\frac{1}{x} - (\frac{x}{x^{2} + 1})] = \frac{x}{\sqrt{x^{2} + 1}} [\frac{1}{x} - (\frac{x}{x^{2} + 1})] = \frac{1}{\sqrt{x^{2} + 1}} - \frac{x^{2}}{{(x^{2} + 1)}^{\frac{3}{2}}}$ .

Page last validated 2010-08-15