We have already calculated some derivatives of elementary functions using the limit of the difference quotient of the function \( f(x) \). Now, we will see how to compute - in a more general way - the derivative of the sum \((f + g )(x_0)\), the derivative of the product \((f \cdot g )(x_0)\), the inverse function \(f^{-1}(x_0)\), and the composite function \((f \circ g)(x_0)\).
Table of Contents
- Derivative of the Sum
- Derivative of the Product
- Derivative of the Composite Function
- Derivative of the Inverse Function
Derivative of the Sum. Let \( f : X \subset \mathbb{R} \to \mathbb{R} \) and \( g : Y \subset \mathbb{R} \to \mathbb{R} \) be two functions, and let \(x_0 \in X\cap Y\). If \( f \) and \( g \) are differentiable at the point \(x_0\), then \( (f + g )(x) \) is differentiable at \(x_0\), and its derivative is given by: \[ (f + g)'(x_0) = f'(x_0) + g'(x_0) \]
Proof. We apply the definition of the derivative to the sum function:
\begin{align} \lim_{x \to x_0} \frac{f(x) + g(x) - (f(x_0) + g(x_0))}{x - x_0} &= \lim_{x \to x_0} \frac{f(x) - f(x_0) + g(x) - g(x_0)}{x - x_0} \\ &= \lim_{x\to x_0}\frac{f(x) - f(x_0)}{x - x_0} + \lim_{x\to x_0}\frac{g(x) - g(x_0)}{x - x_0} \end{align}
The last step is justified by the fact that the limit of the sum is equal to the sum of the limits. Therefore, since the functions \(f\) and \(g\) are differentiable at \(x_0\), the sum is differentiable at \(x_0\):
\[ (f + g)'(x_0) = f'(x_0) + g'(x_0) \]
Derivative of the Product. Let \( f : X \subset \mathbb{R} \to \mathbb{R} \) and \( g : Y \subset \mathbb{R} \to \mathbb{R} \) be two functions, and let \(x_0 \in X \cap Y\). If \( f \) and \( g \) are differentiable at the point \( x_0 \), then the product \( (f \cdot g)(x) \) is differentiable at \(x_0\), and its derivative is given by: \[ (f \cdot g)'(x_0) = f'(x_0) \cdot g(x_0) + f(x_0) \cdot g'(x_0) \]
Proof. We apply the definition of the derivative to the product function:
\[ \lim_{x \to x_0} \frac{f(x)g(x) - f(x_0)g(x_0)}{x - x_0} \]
We can algebraically manipulate the expression by adding and subtracting \( f(x_0)g(x) \) in the numerator to highlight the difference of two terms:
\[ f(x)g(x) - f(x_0)g(x_0) = f(x)g(x) - f(x_0)g(x) + f(x_0)g(x) - f(x_0)g(x_0) \]
We group the terms to factor out common elements:
\[ (f(x) - f(x_0))g(x) + f(x_0)(g(x) - g(x_0)) \]
Now, we can substitute this expression into the limit:
\[ \lim_{x \to x_0} \left( \frac{(f(x) - f(x_0))g(x)}{x - x_0} + \frac{f(x_0)(g(x) - g(x_0))}{x - x_0} \right) \]
We can split this limit into two parts:
\[ \lim_{x \to x_0} \frac{(f(x) - f(x_0))g(x)}{x - x_0} + \lim_{x \to x_0} \frac{f(x_0)(g(x) - g(x_0))}{x - x_0} \]
Consider the first limit:
\[ \lim_{x \to x_0} \frac{(f(x) - f(x_0))g(x)}{x - x_0} = \lim_{x \to x_0} \left( \frac{f(x) - f(x_0)}{x - x_0} \right) g(x_0) = f'(x_0)g(x_0) \]
Since \( \lim_{x \to x_0} g(x) = g(x_0) \), we can factor \( g(x_0) \) out of the limit.
Now, consider the second limit:
\[ \lim_{x \to x_0} \frac{f(x_0)(g(x) - g(x_0))}{x - x_0} = f(x_0) \lim_{x \to x_0} \left( \frac{g(x) - g(x_0)}{x - x_0} \right) = f(x_0)g'(x_0) \]
Combining both results, we get:
\[ (f \cdot g)'(x_0) = f'(x_0)g(x_0) + f(x_0)g'(x_0) \]
This is the product rule for derivatives, which states that the derivative of the product of two functions is given by the sum of the product of the derivative of the first function and the second function, plus the product of the first function and the derivative of the second function.
Derivative of the Composite Function. Let \( f : X \subset \mathbb{R} \to \mathbb{R} \) and \( g : Y \subset \mathbb{R} \to \mathbb{R} \) be two functions, where \( X \) contains a neighborhood of \( x_0 \) and \( Y \) contains a neighborhood of \( g(x_0) \), with \( g(X) \subset Y \). If \( g \) is differentiable at \( x_0 \) and \( f \) is differentiable at \( g(x_0) \), then the composite function \((f \circ g)(x) = f(g(x))\) is differentiable at \( x_0 \), and its derivative is given by:
\[(f \circ g)'(x_0) = f'(g(x_0)) \cdot g'(x_0)\]
Proof. We start from the definition of the derivative as the limit of the difference quotient:
\[ (f \circ g)'(x_0) = \lim_{x \to x_0} \frac{f(g(x)) - f(g(x_0))}{x - x_0}\]
We multiply and divide by \((g(x) - g(x_0))\) to introduce the derivative of \(g\):
\[ = \lim_{x \to x_0} \left( \frac{f(g(x)) - f(g(x_0))}{g(x) - g(x_0)} \cdot \frac{g(x) - g(x_0)}{x - x_0} \right)\]
This expression separates into two limits:
\[ = \lim_{x \to x_0} \frac{f(g(x)) - f(g(x_0))}{g(x) - g(x_0)} \cdot \lim_{x \to x_0} \frac{g(x) - g(x_0)}{x - x_0} \]
The first limit is the derivative of \( f \) at \( g(x_0) \), and the second limit is the derivative of \( g \) at \( x_0 \), so we obtain:
\[ (f \circ g)'(x_0) = f'(g(x_0)) \cdot g'(x_0) \]