Question

Given the function \(f(x)=a x+b,\) find the derivatives of: (a)\(f(x)\) (b) \(x f(x)\) (c) \(1 / f(x)\) \((d) f(x) / x\)

Short answer

(a) \(a\); (b) \(2ax + b\); (c) \(-\frac{a}{(ax+b)^2}\); (d) \(-\frac{b}{x^2}\).

Step-by-step solution

Derivative of a Linear Function

The function given is a linear function of the form \(f(x) = ax + b\). The derivative of a linear function \(y = ax + b\) is the constant \(a\). Therefore, for the given function, the derivative is \(f'(x) = a\).

Derivative of Product Function

We are required to find the derivative of \(x f(x) = x(ax + b)\). This expression simplifies to \(ax^2 + bx\). Use the power rule to differentiate: \(\frac{d}{dx}(ax^2) = 2ax\) and \(\frac{d}{dx}(bx) = b\). Thus, the derivative is \(2ax + b\).

Derivative of Reciprocal Function

For the function \(\frac{1}{f(x)} = \frac{1}{ax + b}\), apply the chain rule. Let \(u = ax + b\), making \(\frac{d}{dx} \left( \frac{1}{u} \right) = -\frac{1}{u^2}\) and \(\frac{du}{dx} = a\). Thus, \(\frac{d}{dx} \left( \frac{1}{f(x)} \right) = -\frac{a}{(ax + b)^2}\).

Derivative of Quotient Function

For the derivative of \(\frac{f(x)}{x} = \frac{ax + b}{x}\), utilize the quotient rule \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\). With \(u = ax + b\) and \(v = x\), \(\frac{du}{dx} = a\) and \(\frac{dv}{dx} = 1\). The derivative is \(\frac{x(a) - (ax + b)(1)}{x^2} = \frac{ax - ax - b}{x^2} = -\frac{b}{x^2}\).

Key concepts

Derivative of a Linear Function

The derivative of a linear function is a simple yet fundamental concept in calculus. A linear function is typically expressed as \( f(x) = ax + b \), where \( a \) and \( b \) are constants. To find the derivative, we essentially look at how the function changes with respect to changes in \( x \).
The beauty of linear functions is that their rate of change is constant. This means regardless of what value \( x \) takes, the derivative will always be the same. Specifically, the derivative of \( f(x) = ax + b \) is simply \( a \).
Why? Because the term \( ax \) changes at a constant rate \( a \), and the constant term \( b \) does not affect the rate of change as it shifts the line vertically. Hence, if you are given a linear function, finding its derivative is as straightforward as identifying the coefficient of \( x \), which is \( a \). This linearity makes such derivatives incredibly predictable and easy to calculate.

Product Rule in Differentiation

The product rule is a vital tool in calculus used to find the derivative of a product of two functions. It comes particularly handy when dealing with expressions like \( x f(x) \).
In symbolic terms, if you have two differentiable functions \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product \( u(x)v(x) \) is:

• \( \frac{d}{dx} [u(x) \cdot v(x)] = u(x) \frac{dv}{dx} + v(x) \frac{du}{dx} \).

Let's apply that here: for \( x f(x) \), we can treat \( u = x \) and \( v = f(x) \), where \( f(x) = ax + b \). Using the product rule, we find:

• The derivative of \( x \) (which is \( u \)) is 1.
• The derivative of \( f(x) \) is \( a \) (as previously discussed).

When these are applied in the formula, it results in
\[ x \cdot a + f(x) \cdot 1 \]. After simplifying, the derivative becomes \( 2ax + b \).
This approach provides a structured method for handling the complexity of product differentiation, turning it into manageable steps.

Chain Rule

When differentiating composite functions, the chain rule is indispensable. It allows us to take derivatives of functions inside other functions, like for the reciprocal function \( \frac{1}{f(x)} \).
The chain rule forms a bridge between the derivative of the outer function and the inner function. In more abstract terms, for two functions \( g(f(x)) \), the derivative is given by:

• \( \frac{d}{dx} [g(f(x))] = g'(f(x)) \cdot f'(x) \).

For the reciprocal function \( g(x) = \frac{1}{x} \), the derivative can be stated as \( g'(x) = -\frac{1}{x^2} \). Applying the chain rule, set
\( u = ax + b \) for \( f(x) \), leading to:

• \( g'(u) = -\frac{1}{u^2} \).
• \( u' = a \), since \( f'(x) = a \).

Therefore, the derivative of \( \frac{1}{f(x)} \) is
\( -\frac{a}{(ax + b)^2} \).
This technique is critical for tackling the derivatives of functions layered within one another, ensuring precision and accuracy in calculus work.

Quotient Rule in Calculus

The quotient rule comes into play when differentiating functions expressed as a quotient or fraction, like \( \frac{f(x)}{x} \). In general, if \( u(x) \) and \( v(x) \) are two differentiable functions, the quotient rule states:

• \( \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \).

For our exercise, take \( u = ax + b \) and \( v = x \), then:

• \( u' = a \).
• \( v' = 1 \).

Applying these into the rule gives:

• \( x \cdot a - (ax + b) \cdot 1 \).
• The expression simplifies to \( ax - ax - b \).

The final derivative thus becomes
\( -\frac{b}{x^2} \).
The quotient rule is a robust approach tailored for handling division of functions, safeguarding calculations against common errors in division-based differentiation.