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 calculate - in a more general way - the derivative of the sum \((f + g )(x_0)\), the derivative of the product \((f \cdot g )(x_0)\), of the inverse function \(f ^{ - 1 }(x_0)\) and 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 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 equals the sum of the limits. We therefore 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 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 \( f(x_0)g(x) \) - to 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 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 let us 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 given by the sum of the product of the derivative of the first function times the second function, plus the product of the first function times 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 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 \((g(x) - g(x_0))\):
\[(f \circ g)'(x_0) = \lim_{x \to x_0} \frac{f(g(x)) - f(g(x_0))}{g(x) - g(x_0)} \cdot \frac{g(x) - g(x_0)}{x - x_0}\]
The limit of the product equals the product of the limits, so we can separate the limit into two parts:
\[(f \circ g)'(x_0) = \lim_{x \to x_0} \frac{f(g(x)) - f(g(x_0))}{g(x) - g(x_0)} \cdot \lim_{x \to x_0} \frac{g(x) - g(x_0)}{x - x_0}\]
Now, we recognize that the first limit is the definition of \(f'(g(x_0))\) and the second limit is the definition of \(g'(x_0)\). Therefore we obtain:
\[(f \circ g)'(x_0) = f'(g(x_0)) \cdot g'(x_0)\]
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 we have: \[ (f^{-1})'(y_0) = \frac{1}{f'(x_0)} \]
Proof. By the chain rule for composite functions: \( ( f ^ { - 1 } \circ f )'( x_0 ) = (f^{-1})' (y_0)\cdot f'(x_0) \). But \( ( f ^ { - 1 } \circ f )'( x_0 ) = 1 \), as it is the identity function on \( X \) and therefore:
\[ (f^{-1})'(y_0) = \frac{1}{f'(x_0)} \]
Remark. If \( f \) is differentiable at \(x_0\) with \(f'(x_0)=0\), then \(f^{-1}\) cannot be differentiable at \( y_0=f(x_0) \) since \( 1/ f'(x_0)\) is undefined.
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 \(f'(0) = 0\).