Derivative of e^x

In this tutorial we shall find the derivative of exponential function $${e^x}$$ and we shall prove the general rules for the differentiation of exponential functions.


Let us suppose that the function is of the form \[y = f\left( x \right) = {e^x}\]


First we take the increment or small change in the function:
\[\begin{gathered} y + \Delta y = {e^{x + \Delta x}} \\ \Delta y = {e^{x + \Delta x}} – y \\ \end{gathered} \]


Putting the value of function $$y = {e^x}$$ in the above equation, we get
\[\Delta y = {e^{x + \Delta x}} – {e^x}\]


Dividing both sides by $$\Delta x$$, we get
\[\frac{{\Delta y}}{{\Delta x}} = \frac{{{e^{x + \Delta x}} – {e^x}}}{{\Delta x}}\]


Taking the limit of both sides as $$\Delta x \to 0$$, we have
\[\begin{gathered} \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^{x + \Delta x}} – {e^x}}}{{\Delta x}} \\ \Rightarrow \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{e^x}\left( {{e^{\Delta x}} – 1} \right)}}{{\Delta x}} \\ \end{gathered} \]


Using the following relation from the limit $$\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} – 1}}{x} = \ln a$$, we have
\[\frac{{dy}}{{dx}} = {e^x}\ln e\]


Now we have the relation $$\ln e = 1$$
\[\frac{d}{{dx}}{e^x} = {e^x}\]


Example: Find the derivative of \[y = f\left( x \right) = {e^{\sin x}}\]

We have the given function as
\[y = {e^{\sin x}}\]

Differentiating with respect to variable $$x$$, we get
\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}{e^{\sin x}}\]

Using the rule, $$\frac{d}{{dx}}{e^x} = {e^x}$$, we get
\[\begin{gathered} \frac{{dy}}{{dx}} = {e^{\sin x}}\frac{d}{{dx}}\left( {\sin x} \right) \\ \Rightarrow \frac{{dy}}{{dx}} = {e^{\sin x}}\cos x \\ \end{gathered} \]