We have already calculated some derivatives of elementary functions using the incremental ratio limit of the function \(f(x)\). Now we will see how to calculate - 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 derivative of the inverse function \(f^{-1}(x_0)\), and the derivative of 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. We can then deduce that 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 so that we can factor out common factors:
\[ (f(x) - f(x_0))g(x) + f(x_0)(g(x) - g(x_0)) \]
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 split this limit into two parts:
\[ \lim_{x \to x_0} \frac{(f(x) - f(x_0))g(x)}{x - x_0} + \lim_{x \to x_0} \frac{f(x_0)(g(x) - g(x_0))}{x - x_0} \]
Let us 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 take \( g(x_0) \) outside 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 obtain:
\[ ( f \cdot g )(x_0) = f'(x_0)g(x_0) + f(x_0)g'(x_0) \]
This is the product rule for derivatives, which states that the derivative of the product of two functions is the sum of the product of the derivative of the first function by the second function, plus the product of the first function by 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 incremental ratio
\[ (f \circ g)'(x_0) = \lim_{x \to x_0} \frac{f(g(x)) - f(g(x_0))}{x - x_0}\]
We multiply and divide by \((g(x) - g(x_0))\):
\[(f \circ g)'(x_0) = \lim_{x \to x_0} \frac{f(g(x)) - f(g(x_0))}{g(x) - g(x_0)} \cdot \frac{g(x) - g(x_0)}{x - x_0}\]
The first part of this expression is the definition of the derivative of \(f\) at \(g(x_0)\), so we get:
\[ f'(g(x_0)) \]
The second part is the definition of the derivative of \(g\) at \(x_0\), so we obtain:
\[ g'(x_0) \]
Combining both, we get the result:
\[ (f \circ g)'(x_0) = f'(g(x_0)) \cdot g'(x_0) \]
Derivative of the Inverse Function
Let \( f : X \subset \mathbb{R} \to Y \subset \mathbb{R} \) be a bijective and continuous function on an open interval \(X\), with inverse \( f^{-1} : Y \to X \). Let \( x_0 \in X \) and let \( y_0 = f(x_0) \). If \(f\) is differentiable at \(x_0\) and \(f'(x_0) \neq 0\), then \(f^{-1}\) is differentiable at \(y_0\) and satisfies: \[ (f^{-1})'(y_0) = \frac{1}{f'(x_0)} \]
Proof. By the differentiation rule for composite functions: \( ( f ^ { - 1 } \circ f )'( x_0 ) = f^{-1} (y_0)\cdot f'(x_0) \). But \( ( f ^ { - 1 } \circ f )'( x_0 ) = 1 \), since it is the identity function on \( X \), therefore:
\[ (f^{-1})'(y_0) = \frac{1}{f'(x_0)} \]
Observation. If \( f \) is differentiable at \(x_0\) with \(f'(x_0)=0\), then \(f^{-1}\) cannot be differentiable at \( y_0=f(x_0) \) because \( 1/ f'(x_0)\) is not defined.
Example. Let \( g : [0, +\infty) \longrightarrow [0, +\infty) \) be defined by \( g(y)=y^{1/3} \). The function \(f^{-1}\) cannot be differentiable at \(y_0=0\) because the inverse \(f(x)=x^3\) is differentiable at \(x_0\) with \(f'(0) = 0\).