The proof of the derivative of the natural exponential \( e^x \) is presented using the limit definition of the derivative. The derivative of a composite function of the form \( e^{u(x)} \) is also presented including examples with their solutions.
The definition of the derivative \( f' \) of a function \( f \) is given by the limit
\[ f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \]
Let \( f(x) = e^x \) and write the derivative of \( e^x\) as follows
\( f'(x) = \lim_{h \to 0} \dfrac{e^{x+h}-e^x}{h} \)
Use the formula \( e^{x+h} = e^x e^h \) to rewrite the derivative of \( e^x \) as
\( f'(x) = \lim_{h \to 0} \dfrac{e^x e^h - e^x}{h} \)
Factor \( e^x \) out in the numerator
\( f'(x) = \lim_{h \to 0} \dfrac{e^x (e^h - 1)}{h} \)
Since \( e^x \) does not depend on \( h \), the above may be rewritten as
\( f'(x) = e^x \lim_{h \to 0} \dfrac{ e^h - 1}{h} \)
We now need to find the limit \( \lim_{h \to 0} \dfrac{ e^h - 1}{h} \).
Let \( y = e^h - 1 \)
and note that
\( \lim_{h \to 0} y = 0\)
We now express h in terms of y
\(e^h = y + 1 \)
take the ln of both sides
\(ln(e^h) = ln(y + 1) \)
simplify the left side using the rule: \( ln(e^x) = x \)
\( h = \ln(y + 1) \)
With the above substitution, we can write
\( \lim_{h \to 0} \dfrac{ e^h - 1}{h} = \lim_{y \to 0} \dfrac{ y}{\ln(y+1)} \)
Rewrite the term on the right as follows
\( = \lim_{y \to 0} \dfrac{ 1}{\dfrac{1}{y}\ln(y+1)} \)
Use power rule of logarithms ( \( a \ln y = \ln y^a \) ) to rewrite the above limit as
\( = \lim_{y \to 0} \dfrac{ 1}{\ln(y+1)^{\dfrac{1}{y}}} \)
Use the limit rule of a quotient and limit of a composite function to rewrite the above as
\( = \dfrac{ \lim_{y \to 0} 1}{\lim_{y \to 0} \ln(y+1)^{\dfrac{1}{y}}} = \dfrac{1}{\ln \left(\lim_{y \to 0} (y+1)^{\dfrac{1}{y}} \right)} \)
One of the definitions of the Euler Constant e is
\( e = \lim_{m \to 0} ( 1 + m) ^{\dfrac{1}{m}} \)
Hence the limit we are looking for is given by
\( \lim_{h \to 0} \dfrac{ e^h - 1}{h} = \dfrac{1}{\ln \left(\lim_{y \to 0} (y+1)^{\dfrac{1}{y}} \right)} = \dfrac{1}{\ln e} = 1\)
which finally gives
\( f'(x) = e^x \lim_{h \to 0} \dfrac{ e^h - 1}{h} = e^x \times 1 = e^x \)
Conclusion: \[ \dfrac{d}{dx} e^x = e^x \]
Note that any function of the form \( f(x) = k e^x \), where k is a constant, is equal to its derivative.
Example 1
Find the derivative of the composite exponential functions
Solution to Example 1