Question

How can I find the derivative of $y = {e^x}$ from first principles?

Answer 1

Hint: We need to plug this function into the definition of the derivative, (1), and do some algebra. First plug the function into the definition of the derivative. Next, make factors of numerator by taking common terms out. Next, take ${e^x}$ out of the limit and use the property of exponential function in limits to find the desired result.

Formula used: Definition of the Derivative from First Principles:
The derivative of $f\left( x \right)$ with respect to $x$ is the function $f'\left( x \right)$ and is defined as,
$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$...........…(1)

Complete step-by-step solution:
First plug the function into the definition of the derivative.
 $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
$ \Rightarrow y'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{x + h}} - {e^x}}}{h}$
$ \Rightarrow y'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^x} \cdot {e^h} - {e^x}}}{h}$
Take ${e^x}$ common in the numerator, we get
$ \Rightarrow y'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^x}\left( {{e^h} - 1} \right)}}{h}$
Now, the ${e^x}$ is not affected by the limit since it doesn’t have any $h$’s in it and so is a constant as far as the limit is concerned. We can therefore factor this out of the limit. This gives,
\[ \Rightarrow y'\left( x \right) = {e^x}\mathop {\lim }\limits_{h \to 0} \dfrac{{{e^h} - 1}}{h}\]
We know that $e$ is the unique positive number for which \[\mathop {\lim }\limits_{h \to 0} \dfrac{{{e^h} - 1}}{h} = 1\].
So, \[y'\left( x \right) = {e^x} \times 1\]
Multiply ${e^x}$ with $1$, we get
\[y'\left( x \right) = {e^x}\]

Therefore, the derivative of $y = {e^x}$ is, \[y'\left( x \right) = {e^x}\].

Additional Information: Here, notice that the limit we’ve got, \[y'\left( x \right) = {e^x}\mathop {\lim }\limits_{h \to 0} \dfrac{{{e^h} - 1}}{h}\] is exactly the definition of the derivative of $f\left( x \right) = {e^x}$ at $x = 0$ , i.e., $f'\left( 0 \right)$. Therefore, the derivative becomes,
$f'\left( x \right) = f'\left( 0 \right){e^x}$

Note: We can also use the power series of exponential function to find the derivative of $y = {e^x}$.
Power Series of Exponential function:
${e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ...$
Now, differentiate it with respect to $x$.
$\dfrac{d}{{dx}}\left( {{e^x}} \right) = \dfrac{d}{{dx}}\left( {1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ...} \right)$
Now, use property $\dfrac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$.
$ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}} \right) = \dfrac{d}{{dx}}\left( 1 \right) + \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{2!}}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{x^3}}}{{3!}}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{x^4}}}{{4!}}} \right) + ...$
Now, use property $\dfrac{d}{{dx}}\left[ {kf\left( x \right)} \right] = k\dfrac{d}{{dx}}f\left( x \right)$.
$ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}} \right) = \dfrac{d}{{dx}}\left( 1 \right) + \dfrac{d}{{dx}}\left( x \right) + \dfrac{1}{{2!}}\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{1}{{3!}}\dfrac{d}{{dx}}\left( {{x^3}} \right) + \dfrac{1}{{4!}}\dfrac{d}{{dx}}\left( {{x^4}} \right) + ...$
Now, use property $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$.
$ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}} \right) = 0 + 1 + \dfrac{1}{2} \times 2x + \dfrac{1}{{3 \times 2!}} \times 3{x^2} + \dfrac{1}{{4 \times 3!}} \times 4{x^3} + ...$
$ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}} \right) = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + ...$
$ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$
Therefore, the derivative of $y = {e^x}$ is, \[y'\left( x \right) = {e^x}\].