Derivative of the Sum, Difference, Product, and Quotient of Functions
To compute the derivative of a sum, difference, product, or quotient of functions, we apply the following fundamental differentiation rules:
$$ (f+g)'=f'+g' $$ $$ (f-g)'=f'-g' $$ $$ (f \cdot g)'=f' \cdot g + f \cdot g' $$ $$ \left( \frac{f}{g} \right)'= \frac{f' \cdot g - f \cdot g'}{g^2} $$ $$ (k \cdot f)' = k \cdot f' $$
Each of these rules can be rigorously derived by evaluating the limit definition of the derivative for the corresponding operations.
Derivative of the Sum of Functions
The derivative of the sum of two functions, \( (f+g)' \), is simply the sum of their derivatives: \( f' + g' \).
$$ (f+g)'=f'+g' $$
Proof
We begin by computing the difference quotient for \( f+g \):
$$ \lim_{h \rightarrow 0} \frac{[f(x+h)+g(x+h)] - [f(x)+g(x)]}{h} $$
Expanding the numerator yields:
$$ \lim_{h \rightarrow 0} \left( \frac{f(x+h) - f(x)}{h} + \frac{g(x+h) - g(x)}{h} \right) $$
Applying the limit separately to each term, we obtain:
$$ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} + \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h} $$
where:
$$ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = f'(x) \quad \text{and} \quad \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h} = g'(x) $$
Thus,
$$ (f+g)' = f' + g' $$
This confirms that the derivative of a sum is the sum of the derivatives.
Derivative of the Difference of Functions
Similarly, the derivative of the difference of two functions, \( (f-g)' \), is the difference of their derivatives: \( f' - g' \).
$$ (f-g)'=f'-g' $$
Proof
Computing the difference quotient for \( f-g \) gives:
$$ \lim_{h \rightarrow 0} \frac{[f(x+h)-g(x+h)] - [f(x)-g(x)]}{h} $$
Expanding and rearranging the terms, we have:
$$ \lim_{h \rightarrow 0} \left( \frac{f(x+h)-f(x)}{h} - \frac{g(x+h)-g(x)}{h} \right) $$
Taking the limit of each term separately:
$$ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} - \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h} $$
thus:
$$ (f-g)' = f' - g' $$
Hence, the derivative of a difference is the difference of the derivatives.
Derivative of the Product of Functions
The derivative of the product of two functions, \( (f \cdot g)' \), follows the product rule: it is the sum of the product of the first function and the derivative of the second plus the product of the second function and the derivative of the first.
$$ (f \cdot g)'=f' \cdot g + f \cdot g' $$
Proof
Starting with the difference quotient for \( f \cdot g \):
$$ \lim_{h \rightarrow 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h} $$
We introduce and subtract the term \( f(x+h)g(x) \) to rearrange the expression:
$$ \lim_{h \rightarrow 0} \frac{[f(x+h)g(x+h) - f(x+h)g(x)] + [f(x+h)g(x) - f(x)g(x)]}{h} $$
Grouping terms gives:
$$ \lim_{h \rightarrow 0} \left( f(x+h) \frac{g(x+h) - g(x)}{h} + g(x) \frac{f(x+h) - f(x)}{h} \right) $$
Taking the limit as \( h \to 0 \), \( f(x+h) \to f(x) \) and we find:
$$ f(x) \cdot g'(x) + g(x) \cdot f'(x) $$
Thus,
$$ (f \cdot g)' = f' \cdot g + f \cdot g' $$
This establishes the product rule for derivatives.
Derivative of the Product of a Function and a Constant
The derivative of the product of a function \( f \) and a constant \( k \) is simply the constant multiplied by the derivative of the function:
$$ (k \cdot f)' = k \cdot f' $$
In other words, when differentiating the product of a constant and a function, the constant can be factored outside the derivative operator.
Proof
This result follows directly from the product rule. Here, one of the "functions" involved is the constant \( k \).
Applying the product rule:
$$ (k \cdot f)' = k' \cdot f + k \cdot f' $$
Since the derivative of a constant is zero, \( k' = 0 \), we obtain:
$$ (k \cdot f)' = 0 \cdot f + k \cdot f' = k \cdot f' $$
This completes the proof: the derivative of the product of a constant and a function is the constant times the derivative of the function.
Derivative of the Quotient of Functions
The derivative of the quotient of two functions, \( (f/g)' \), is given by the difference of the cross-products divided by the square of the denominator:
$$ \left( \frac{f}{g} \right)' = \frac{f' \cdot g - f \cdot g'}{g^2} $$
Proof
We start by rewriting the quotient as the product of \( f(x) \) and the reciprocal of \( g(x) \):
$$ \frac{f}{g} = f \cdot \frac{1}{g} $$
We then differentiate each factor individually.
Derivative of \( f \)
By definition:
$$ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f'(x) $$
Derivative of \( 1/g \)
We compute the difference quotient for \( 1/g \):
$$ \lim_{h \to 0} \frac{ \frac{1}{g(x+h)} - \frac{1}{g(x)} }{h} $$
Combining the terms in the numerator yields:
$$ \lim_{h \to 0} \left( \frac{g(x) - g(x+h)}{g(x+h) \cdot g(x)} \right) \cdot \frac{1}{h} $$
which can be rewritten as:
$$ \lim_{h \to 0} -\left( \frac{g(x+h) - g(x)}{h} \right) \cdot \frac{1}{g(x+h) \cdot g(x)} $$
Taking the limit separately, we have:
$$ -g'(x) \cdot \lim_{h \to 0} \frac{1}{g(x+h) \cdot g(x)} $$
Since \( g(x+h) \to g(x) \) as \( h \to 0 \), it follows that:
$$ -g'(x) \cdot \frac{1}{g(x)^2} $$
thus:
$$ \frac{-g'(x)}{g(x)^2} $$
Derivative of the quotient \( f/g \)
Applying the product rule to \( f(x) \cdot \frac{1}{g(x)} \) gives:
$$ \left( f \cdot \frac{1}{g} \right)' = f' \cdot \frac{1}{g} + f \cdot \left( \frac{1}{g} \right)' $$
Substituting the expressions computed above:
$$ = f' \cdot \frac{1}{g} + f \cdot \left( \frac{-g'}{g^2} \right) $$
which simplifies to:
$$ = \frac{f'}{g} - \frac{f \cdot g'}{g^2} $$
Combining the two terms over a common denominator:
$$ = \frac{f' \cdot g - f \cdot g'}{g^2} $$
This completes the proof of the quotient rule.
