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: \( f' + g' \).

$$ (f+g)'=f'+g' $$

Example

Let's find the derivative of the function $ y=x^2 + 3x $.

$$ y' = D[x^2+3x] $$

Using the sum rule, we can differentiate each term separately:

$$ y' = D[x^2]+ D[3x] $$

Since $ D[x^2]=2x $ and $ D[3x]= 3 $, we obtain:

$$ y' = 2x+ 3 $$

This approach works for any sum of functions, even when more than two terms are involved.

Proof

We begin by computing the difference quotient for \( f+g \):

$$ \lim_{h \rightarrow 0} \frac{[f(x+h)+g(x+h)] - [f(x)+g(x)]}{h} $$

$$ \lim_{h \rightarrow 0} \frac{f(x+h)+g(x+h) - f(x)-g(x)}{h} $$

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: \( f' - g' \).

$$ (f-g)'=f'-g' $$

Example

Now let's find the derivative of the function $ y=x^2 - 3x $.

$$ y' = D[x^2-3x] $$

The same rule applies when the terms are being subtracted:

$$ y' = D[x^2]- D[3x] $$

Since $ D[x^2]=2x $ and $ D[3x]= 3 $, we get:

$$ y' = 2x - 3 $$

This is the derivative of the function.

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.

$$ (f \cdot g)'=f' \cdot g + f \cdot g' $$

Example

Let's find the derivative of the function $ y = x^2 \cdot \sin x $.

$$ y' = D[ x^2 \cdot \sin x ] $$

Because the function is the product of two functions, $ x^2 $ and $ \sin x $, we use the product rule:

$$ y' = D[ x^2 ] \cdot \sin x + x^2 \cdot D[ \sin x ] $$

Next, compute the derivative of each function:

$$ D[x^2] = 2x $$

$$ D[\sin x] = \cos x $$

Substituting these results into the formula gives:

$$ y' = 2x \cdot \sin x + x^2 \cdot \cos x $$

Therefore, the derivative of the function is:

$$ y' = 2x \sin x + x^2 \cos x $$

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.

Example

Let's find the derivative of the function $ y= 3 x^2 $.

$$ y'=D[3 \cdot x^2] $$

According to the constant multiple rule, the constant can be moved outside the differentiation operator:

$$ y'=3 \cdot D[x^2] $$

Since $ D[x^2] = 2x $, we have:

$$ y'=3 \cdot 2x $$

$$ y' = 6x $$

This is the derivative of the original function.

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.

Alternative Proof

Another way to prove the constant multiple rule is to start directly from the definition of the derivative and evaluate the difference quotient for the function \( k \cdot f(x) \).

$$ \lim_{h \rightarrow 0} \frac{  k \cdot f(x+h) - k \cdot f(x)  }{h} $$

Since the constant \( k \) appears in both terms of the numerator, it can be factored out.

$$ \lim_{h \rightarrow 0} \frac{  k \cdot \left[ f(x+h) - f(x) \right]  }{h} $$

Because \( k \) does not depend on \( h \), it can be moved outside the limit.

$$ k \cdot \lim_{h \rightarrow 0} \frac{  f(x+h) - f(x)  }{h} $$

The remaining limit is exactly the definition of the derivative of \( f(x) \).

$$ k \cdot f'(x) $$

We have therefore shown that differentiating a function multiplied by a constant is equivalent to multiplying the derivative of the function by that same constant.

This establishes the constant multiple rule.

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.