Derivative of \({e}^x\) using First Principle

Derivative of \({e}^x\) using the First Principle

Let \(y\) = \({e}^x\)
∴ \(y + δy\) = \({e}^{x + δx}\)
∴ \(δy\) = \({e}^{x + δx}\) – \({e}^x\)
or \(δy\) = \({e}^{x}\) . \( [ {e}^{δx} – 1 ]\)
Dividing each side by δx </h3>
or \(\dfrac {δy}{δx}\) = \( \dfrac { {e}^{x} . [ {e}^{δx} – 1 ] } {δx}\)

∴ \(\dfrac {dy}{dx} = \) \( \lim_{δx \to 0} \) \( \dfrac { {e}^{x} . [ {e}^{δx} – 1 ] } {δx}\)
or \(\dfrac {dy}{dx}\) = \( {e}^{x} .\) \( \lim_{δx \to 0} \) \( \dfrac { [ {e}^{δx} – 1 ] } {δx}\) —– (1)

Now, \( \lim_{δx \to 0} \) \( \dfrac { [ {e}^{δx} – 1 ] } {δx}\) =
\( \lim_{δx \to 0} \) \( \dfrac { [ 1 + \dfrac {{δx}}{1!} + \dfrac {{δx}^{2}}{2!} + \dfrac {{δx}^{3}}{3!} +\ … ] – 1} {δx}\)

= \( \lim_{δx \to 0} \) \( \dfrac { \dfrac {{δx}}{1!} + \dfrac {{δx}^{2}}{2!} + \dfrac {{δx}^{3}}{3!} +\ … } {δx}\)

Cancelling δx in both numerator and denominator, we get

\( \lim_{δx \to 0} \) \( \dfrac { [ {e}^{δx} – 1 ] } {δx}\) = \( \lim_{δx \to 0} \) \( \left( \dfrac {1}{1!} + \dfrac {{δx}^{1}}{2!} + \dfrac {{δx}^{2}}{3!} +\ … \right) \)

or \( \lim_{δx \to 0} \) \( \dfrac { [ {e}^{δx} – 1 ] } {δx}\) = \( \left( \dfrac {1}{1!} + \dfrac {{0}^{1}}{2!} + \dfrac {{0}^{2}}{3!} +\ … \right) \) = 1 —– (2)

∴ from equations (1) and (2), we get
\(\dfrac {dy}{dx} \) = \( {e}^{x} .\) 1 = \( {e}^{x} \)