We have already calculated some derivatives of elementary functions using the limit of the difference quotient. We now study, 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 composite function \((f\circ g)'(x_0)\), and the derivative of the inverse function \((f^{-1})'(y_0)\).
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:
\[ \lim_{x \to x_0} \frac{(f(x) - f(x_0))g(x)}{x - x_0} = \lim_{x \to x_0} \left( \frac{f(x)-f(x_0)}{x-x_0}\cdot g(x) \right). \]
Since \(g\) is differentiable at \(x_0\), it is continuous at \(x_0\). Therefore
\[ \lim_{x \to x_0}g(x)=g(x_0). \]
Hence
\[ \lim_{x \to x_0} \frac{(f(x) - f(x_0))g(x)}{x - x_0} = f'(x_0)g(x_0). \]
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 \(g:X\subset\mathbb{R}\to\mathbb{R}\) and \(f:Y\subset\mathbb{R}\to\mathbb{R}\) be two functions such that \(g(X)\subset Y\). Let \(x_0\in X\), and suppose that \(g\) is differentiable at \(x_0\) and that \(f\) is differentiable at \(g(x_0)\). Then the composite function
\[ (f\circ g)(x)=f(g(x)) \]
is differentiable at \(x_0\), and
\[ (f\circ g)'(x_0)=f'(g(x_0))\cdot g'(x_0). \]
Proof. Set
\[ y_0=g(x_0). \]
Since \(f\) is differentiable at \(y_0\), there exists a function \(\varepsilon\), with \(\varepsilon(y)\to 0\) as \(y\to y_0\), such that
\[ f(y)-f(y_0)=\bigl(f'(y_0)+\varepsilon(y)\bigr)(y-y_0). \]
Applying this identity with \(y=g(x)\), we obtain
\[ f(g(x))-f(g(x_0)) = \bigl(f'(g(x_0))+\varepsilon(g(x))\bigr) \bigl(g(x)-g(x_0)\bigr). \]
Dividing by \(x-x_0\), with \(x\neq x_0\), gives
\[ \frac{f(g(x))-f(g(x_0))}{x-x_0} = \bigl(f'(g(x_0))+\varepsilon(g(x))\bigr) \frac{g(x)-g(x_0)}{x-x_0}. \]
Since \(g\) is differentiable at \(x_0\), it is continuous at \(x_0\). Hence, as \(x\to x_0\), we have \(g(x)\to g(x_0)\), and therefore
\[ \varepsilon(g(x))\to 0. \]
Passing to the limit, we obtain
\[ (f\circ g)'(x_0) = f'(g(x_0))\cdot g'(x_0). \]
This proves the chain rule.
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 set
\[ y_0=f(x_0). \]
If \(f\) is differentiable at \(x_0\) and \(f'(x_0)\neq 0\), then \(f^{-1}\) is differentiable at \(y_0\), and
\[ (f^{-1})'(y_0)=\frac{1}{f'(x_0)}. \]
Proof. Consider the difference quotient of the inverse function:
\[ \frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}. \]
Set
\[ x=f^{-1}(y). \]
Then \(y=f(x)\). Moreover, since \(f^{-1}\) is continuous, the condition \(y\to y_0\) implies \(x\to x_0\). We can therefore write
\[ \frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0} = \frac{x-x_0}{f(x)-f(x_0)}. \]
Passing to the limit as \(y\to y_0\), that is, as \(x\to x_0\), we obtain
\[ \lim_{y\to y_0} \frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0} = \lim_{x\to x_0} \frac{x-x_0}{f(x)-f(x_0)}. \]
Since \(f'(x_0)\neq 0\), we have
\[ \lim_{x\to x_0} \frac{x-x_0}{f(x)-f(x_0)} = \frac{1}{\displaystyle\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}} = \frac{1}{f'(x_0)}. \]
Hence
\[ (f^{-1})'(y_0)=\frac{1}{f'(x_0)}. \]
Remark. If \(f'(x_0)=0\), the formula above cannot be applied. Indeed, if \(f^{-1}\) were differentiable at \(y_0=f(x_0)\), then applying the chain rule to the identity
\[ (f^{-1}\circ f)(x)=x \]
would give
\[ (f^{-1})'(y_0)\cdot f'(x_0)=1, \]
which is impossible when \(f'(x_0)=0\).
Example. Consider the function \(f:\mathbb{R}\to\mathbb{R}\) defined by
\[ f(x)=x^3. \]
It is bijective and has inverse
\[ f^{-1}(y)=\sqrt[3]{y}. \]
However, \(f'(0)=0\). Indeed, the inverse function is not differentiable at \(y_0=0\), because
\[ \frac{f^{-1}(y)-f^{-1}(0)}{y-0} = \frac{\sqrt[3]{y}}{y} = \frac{1}{\sqrt[3]{y^2}} \]
does not have a finite limit as \(y\to 0\).