Derivative of Algebraic Functions

Algebraic functions are functions that can be expressed using polynomial expressions, rational expressions (quotients of polynomials), and roots. Differentiation is a fundamental concept in calculus that deals with the rate at which a function is changing at any given point. The derivative of a function at a point is the slope of the tangent line to the graph of the function at that point.

Basic Derivative Rules

Before we delve into the derivatives of algebraic functions, let's review some basic derivative rules:

Rule	Formula	Example
Constant Rule	$\frac{d}{dx}(c) = 0$	$\frac{d}{dx}(7) = 0$
Power Rule	$\frac{d}{dx}(x^n) = nx^{n-1}$	$\frac{d}{dx}(x^3) = 3x^2$
Constant Multiple Rule	$\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))$	$\frac{d}{dx}(5x^2) = 5\frac{d}{dx}(x^2) = 10x$
Sum Rule	$\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))$	$\frac{d}{dx}(x^2 + 3x) = 2x + 3$
Difference Rule	$\frac{d}{dx}(f(x) - g(x)) = \frac{d}{dx}(f(x)) - \frac{d}{dx}(g(x))$	$\frac{d}{dx}(x^3 - x) = 3x^2 - 1$
Product Rule	$\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$	$\frac{d}{dx}(x^2 \cdot x^3) = 2x \cdot x^3 + x^2 \cdot 3x^2$
Quotient Rule	$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$	$\frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{2x(x+1) - x^2}{(x+1)^2}$

Derivative of Polynomial Functions

Polynomial functions are of the form $P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, where $a_n, a_{n-1}, \ldots, a_1, a_0$ are constants. The derivative of a polynomial function is found by applying the power rule to each term.

Example:

Find the derivative of $P(x) = 2x^3 - 5x^2 + 3x - 7$.

Solution:

$$ P'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(5x^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(7) $$

Applying the power rule:

$$ P'(x) = 2 \cdot 3x^{3-1} - 5 \cdot 2x^{2-1} + 3 \cdot 1x^{1-1} - 0 $$

Simplifying:

$$ P'(x) = 6x^2 - 10x + 3 $$

Derivative of Rational Functions

Rational functions are of the form $R(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomial functions. The derivative of a rational function is found by applying the quotient rule.

Example:

Find the derivative of $R(x) = \frac{x^2 - 1}{x + 2}$.

Solution:

Let $P(x) = x^2 - 1$ and $Q(x) = x + 2$. Then $P'(x) = 2x$ and $Q'(x) = 1$. Applying the quotient rule:

$$ R'(x) = \frac{P'(x)Q(x) - P(x)Q'(x)}{Q(x)^2} $$

Substituting the derivatives and simplifying:

$$ R'(x) = \frac{2x(x + 2) - (x^2 - 1)(1)}{(x + 2)^2} $$

Expanding and simplifying:

$$ R'(x) = \frac{2x^2 + 4x - x^2 + 1}{(x + 2)^2} $$

Combining like terms:

$$ R'(x) = \frac{x^2 + 4x + 1}{(x + 2)^2} $$

Derivative of Root Functions

Root functions are of the form $f(x) = \sqrt[n]{g(x)}$, where $g(x)$ is a polynomial function. To differentiate root functions, we can rewrite them as $f(x) = g(x)^{\frac{1}{n}}$ and apply the power rule.

Example:

Find the derivative of $f(x) = \sqrt[3]{x^2}$.

Solution:

Rewrite the function as $f(x) = (x^2)^{\frac{1}{3}}$. Then apply the power rule:

$$ f'(x) = \frac{1}{3}(x^2)^{\frac{1}{3} - 1} \cdot 2x $$

Simplifying:

$$ f'(x) = \frac{2}{3}x^{\frac{-1}{3}} $$

Conclusion

Understanding the derivatives of algebraic functions is crucial for solving problems in calculus. By applying the basic rules of differentiation, such as the power rule, product rule, and quotient rule, we can find the derivatives of polynomial, rational, and root functions. Practice with various examples will help solidify these concepts and prepare you for exams.