Question

Find the derivatives of: \((a)\left(x^{2}+3\right) / x\) \((b)(x+9) / x\) (c) \(6 x /(x+5)\) \((d)\left(a x^{2}+b\right) /(c x+d)\)

Short answer

(a) \( 1 - \frac{3}{x^2} \), (b) \( \frac{-9}{x^2} \), (c) \( \frac{30}{(x+5)^2} \), (d) \( \frac{acx^2 + 2adx - bc}{(cx+d)^2} \).

Step-by-step solution

Understanding the Quotient Rule

The quotient rule is used when differentiating a function that is the quotient of two functions. If you have a function of the form \( \frac{u(x)}{v(x)} \), the derivative \( \frac{d}{dx} \left( \frac{u}{v} \right) \) is given by: \[ \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \]Let's use this rule to differentiate each given function.

Differentiate Part (a)

For the function \( f(x) = \frac{x^2 + 3}{x} \), identify \( u(x) = x^2 + 3 \) and \( v(x) = x \).Calculate the derivatives:- \( u'(x) = 2x \)- \( v'(x) = 1 \)Apply the quotient rule:\[\frac{d}{dx} \left( \frac{x^2 + 3}{x} \right) = \frac{x \cdot 2x - (x^2 + 3) \cdot 1}{x^2} \]Simplify:\[\frac{2x^2 - x^2 - 3}{x^2} = \frac{x^2 - 3}{x^2} = 1 - \frac{3}{x^2}\]

Differentiate Part (b)

For the function \( g(x) = \frac{x + 9}{x} \), let \( u(x) = x + 9 \) and \( v(x) = x \).Derivatives are:- \( u'(x) = 1 \)- \( v'(x) = 1 \)Using the quotient rule:\[\frac{d}{dx} \left( \frac{x + 9}{x} \right) = \frac{x \cdot 1 - (x + 9) \cdot 1}{x^2} \]Simplify:\[\frac{x - x - 9}{x^2} = \frac{-9}{x^2}\]

Differentiate Part (c)

For the function \( h(x) = \frac{6x}{x + 5} \), choose \( u(x) = 6x \) and \( v(x) = x + 5 \).Derivatives are:- \( u'(x) = 6 \)- \( v'(x) = 1 \)Apply the quotient rule:\[\frac{d}{dx} \left( \frac{6x}{x + 5} \right) = \frac{(x+5) \cdot 6 - 6x \cdot 1}{(x+5)^2} \]Simplify:\[\frac{6x + 30 - 6x}{(x+5)^2} = \frac{30}{(x+5)^2}\]

Differentiate Part (d)

For the function \( i(x) = \frac{ax^2 + b}{cx + d} \), identify \( u(x) = ax^2 + b \) and \( v(x) = cx + d \).Calculate the derivatives:- \( u'(x) = 2ax \)- \( v'(x) = c \)Use the quotient rule:\[\frac{d}{dx} \left( \frac{ax^2 + b}{cx + d} \right) = \frac{(cx + d) \cdot 2ax - (ax^2 + b) \cdot c}{(cx + d)^2} \]Simplify:\[\frac{2acx^2 + 2adx - acx^2 - bc}{(cx+d)^2} = \frac{acx^2 + 2adx - bc}{(cx+d)^2}\]

Conclusion

By applying the quotient rule to each function, we have found the derivatives:(a) \( 1 - \frac{3}{x^2} \), (b) \( \frac{-9}{x^2} \), (c) \( \frac{30}{(x+5)^2} \), and (d) \( \frac{acx^2 + 2adx - bc}{(cx+d)^2} \).

Key concepts

Quotient Rule

When dealing with derivatives of functions that are quotients of other functions, it's essential to use the *quotient rule*. This rule is a crucial differentiation technique that specifically applies to functions in the form \( \frac{u(x)}{v(x)} \). The formula for the derivative in this case is: \[ \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \] Here is how it works:

• Identify the functions: In the expression \( \frac{u}{v} \), \( u(x) \) is the numerator and \( v(x) \) is the denominator.
• Find the derivatives: Determine \( u'(x) \) and \( v'(x) \), the derivatives of \( u(x) \) and \( v(x) \).
• Apply the quotient rule: Substitute into the formula. Be meticulous with signs as errors here can lead to incorrect results.

The quotient rule helps in breaking down complex differentiation into manageable steps by methodically handling the two components of the quotient function.

Differentiation Techniques

Differentiation techniques encompass various methods used to find derivatives, key among them being the quotient rule we've discussed. To efficiently work through problems involving derivatives, it's crucial to become familiar with several standard techniques:

• Power Rule: The simplest method, allowing you to differentiate \( x^n \) as \( nx^{n-1} \).
• Product Rule: Used when differentiating the product of two functions, given by \( (uv)' = u'v + uv' \).
• Chain Rule: Essential for functions within functions, such as \( f(g(x)) \). This is given by \( f'(g(x)) \cdot g'(x) \).
• Quotient Rule: Specialized for quotients as discussed, especially useful for functions expressed as \( \frac{u(x)}{v(x)} \).

By understanding and practicing these techniques, you'll improve your ability to differentiate a wide variety of functions accurately. Each technique has its specific scenarios where it's best applied, so becoming adept at identifying these situations is key.

Mathematical Functions

Mathematical functions, often expressed as \( f(x) \), serve as the backbone of calculus and differentiation. Understanding these functions is critical to applying differentiation techniques effectively.

• Basic Definitions: A function is a relation that associates an input (\( x \)) to a single output (\( f(x) \)). Functions can be linear, quadratic, polynomial, exponential, etc.
• Polynomial Functions: These consist of terms like \( ax^n \), where \( a \) is a constant and \( n \) is a non-negative integer. They're particularly notable for their fixed degree.
• Rational Functions: Functions like \( \frac{p(x)}{q(x)} \) where both \( p \) and \( q \) are polynomials. Here the quotient rule is often used for differentiation.
• Piecewise Functions: Defined by different expressions based on intervals of \( x \), requiring special attention during differentiation at the boundaries.

Comprehending these types of functions allows you to properly choose and apply differentiation rules, such as the quotient rule, when solving calculus problems. Mastery of function types enhances problem-solving skills and makes mathematical analysis easier.