Proof of Derivative of \( e^x \)

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.

Proof of the Derivative of \( e^x \) Using the Definition of the Derivative

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.

Derivative of the Composite Function \( y = e^{u(x)} \)

Example 1
Find the derivative of the composite exponential functions

1. \( f(x) = e^{x^3-2x+3} \)
2. \( g(x) = e^{\sqrt{x^2+1}} \)
3. \( h(x) = e^{ \left(\dfrac{x}{x-2}\right)} \)

Solution to Example 1

1. 
   Let \( u(x) = x^3-2x+3 \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} (x^3-2x+3) = 3x^2-2 \)
   Apply the rule for the composite exponential function found above
   \( \displaystyle \dfrac{d}{dx} f(x) = e^u \dfrac{d}{dx} u = e^{x^3-2x+3} \times (3x^2-2) \)
   \( = (3x^2-2) e^{x^3-2x+3} \)
   
   
2. 
   Let \( u(x) = \sqrt{x^2+1} \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} \sqrt{x^2+1} = \dfrac{x}{\sqrt{x^2+1}} \).
   Apply the above rule of differentiation for the composite exponential function
   \( \displaystyle \dfrac{d}{dx} g(x) = e^u \dfrac{d}{dx} u = e^{\sqrt{x^2+1}} \times \dfrac{x}{\sqrt{x^2+1}} \)
   \( = \dfrac{x}{\sqrt{x^2+1}} \; e^{\sqrt{x^2+1}} \)
   
   
3. 
   Let \( u(x) = \dfrac{x}{x-2} \) and therefore \( \dfrac{d}{dx} u = -\dfrac{2}{\left(x-2\right)^2} \)
   Apply the rule of differentiation for the composite exponential function obtained above
   \( \displaystyle \dfrac{d}{dx} h(x) = e^u \dfrac{d}{dx} u = e^{ \left(\dfrac{x}{x-2}\right)} \times ( -\dfrac{2}{\left(x-2\right)^2} ) \)
   \( = -\dfrac{2}{\left(x-2\right)^2} \; e^{\left(\dfrac{x}{x-2}\right)} \)
   
   

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