Derivada de cociente de funciones

How to derive quotients of functions?

Mastering the rules of differentiation is crucial for those interested in differential calculus. This time we will focus on the derivative of the quotient of functions, a rule similar to that of the product, but with its own characteristics and details that ensure its correct use.

What is the quotient rule?

The quotient rule is applied when we have a fraction of functions, ( Òfrac{f(x)}{g(x)} ), and we want to find its derivative. The formula is:

[ [ \text{derivative} = \frac{derivative of numerator} - \text{numerator}{derivative of denominator}{(\text{denominator})^2} ] ].

Compared to the product rule, the difference is that here a subtraction is used instead of an addition. This marks the big distinction to remember: we multiply, subtract, and normalize the denominator to the square.

Practical examples of quotient derivation

Example 1: Simple Binomials

Consider the function (y = \frac{x - 1}{3x + 2}).

1. Derive the numerator: The derivative of (x - 1) is 1.
2. Multiply by the denominator without altering: You have (3x + 2).
3. Subtract the product of the numerator by the derivative of the denominator: The derivative of (3x) is 3, while the constant 2 disappears.
4. Normalize: The denominator is squared, ((3x + 2)^2).

Operations:

• (1 \times (3x + 2) = 3x + 2).
• Subtract what is obtained from the previous calculation adjusting the negative sign: (-[(x - 1) \times 3] = -[3x - 3]).

Simplified result: [ \frac{5}{(3x + 2)^2} ].

Example 2: Terms of higher complexity.

Take the function ( y = \frac{5x^3 - 3x^2}{-2x + 6} ).

1. Derive the numerator:

   • (5x^3 \mapsto 15x^2)
   • (-3x^2 \mapsto -6x)

2. Multiply by the denominator:

   • Persist (-2x + 6).

3. Inverse subtraction:

   • Derive (-2x + 6 \mapsto -2).
   • Multiply by the numerator without changing: ( 5x^3 - 3x^2).

4. Normalize: Denominator squared, ((-2x + 6)^2).

It is left as a task to unravel the pending algebra and confirm using educational resources.

Example 3: Powers and Roots

Let's look at ( y = \frac{\sqrt{x}}{x^2 + 1} ).

1. Restatement: (\sqrt{x} = x^{1/2})

2. Deriving the numerator: Derivative of ( x^{1/2} ) is ( \frac{1}{2sqrt{x}} ).

3. Multiplying by the initial denominator: The denominator function remains the same.

4. Completing the denominator squared:

   • Derivative of (x^2 + 1) is (2x). Multiplying this by the numerator.

Complexities in complex fractions demand detailed algebraic work. Students are encouraged to continue exploring the supports.

Why practicing is essential?

We have acquired various derivation tools: constants, products, quotients.... However, the key is to reinforce them with constant practice. The proposed assignment, "Multiplying and dividing when deriving", offers a range of exercises that will consolidate both your skills and your confidence in differential calculus. Don't be discouraged and keep learning!