Differentiate e²x with respect to x from first-principles

To differentiate the function \( e^{2x} \) with respect to \( x \) using first principles, we start with the definition of the derivative.

Definition of the Derivative

The derivative of a function \( f(x) \) at a point \( x \) is given by:

\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

Applying the Definition

For our function \( f(x) = e^{2x} \), we need to calculate \( f(x+h) \):

• Calculate \( f(x+h) = e^{2(x+h)} = e^{2x + 2h} = e^{2x} \cdot e^{2h} \)

Now, substitute \( f(x) \) and \( f(x+h) \) into the derivative formula:

\( f'(x) = \lim_{h \to 0} \frac{e^{2x} \cdot e^{2h} - e^{2x}}{h} \)

Simplifying the Expression

Factor out \( e^{2x} \):

\( f'(x) = e^{2x} \cdot \lim_{h \to 0} \frac{e^{2h} - 1}{h} \)

Evaluating the Limit

The limit \( \lim_{h \to 0} \frac{e^{2h} - 1}{h} \) can be evaluated using the fact that \( \frac{e^u - 1}{u} \) approaches 1 as \( u \) approaches 0. Here, let \( u = 2h \), so as \( h \) approaches 0, \( u \) also approaches 0:

• Thus, \( \lim_{h \to 0} \frac{e^{2h} - 1}{h} = 2 \cdot 1 = 2 \)

Final Result

Now, substituting back into our expression for the derivative:

\( f'(x) = e^{2x} \cdot 2 = 2e^{2x} \)

Therefore, the derivative of \( e^{2x} \) with respect to \( x \) is:

\( \frac{d}{dx}(e^{2x}) = 2e^{2x} \)