# Differentiation of Exponential Function

Here you will learn differentiation of exponential function by using first principle and its examples.

Let’s begin –

## Differentiation of Exponential Function

### (1) Differentiation of $$e^x$$ :

The differentiation of $$e^x$$ with respect to x is $$e^x$$.

i.e. $$d\over dx$$ $$e^x$$ = $$e^x$$

#### Proof Using first Principle :

Let f(x) = $$e^x$$. Then, f(x + h) = $$e^{x + h}$$

$$\therefore$$   $$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$f(x + h) – f(x)\over h$$

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$e^{x + h} – e^x\over h$$

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$e^x.e^h – e^x\over h$$

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$e^x$$ ($$e^h – 1\over h$$)

$$\implies$$ $$d\over dx$$(f(x)) = $$e^x$$  $$lim_{h\to 0}$$ ($$e^h – 1\over h$$)

because, [$$lim_{h\to 0}$$($$e^h – 1\over h$$) = 1]

$$\implies$$ $$d\over dx$$(f(x)) = $$e^x$$ $$\times$$ 1 = $$e^x$$

Hence, $$d\over dx$$ ($$e^x$$) = $$e^x$$

Example : What is the differentiation of $$e^{2x}$$ ?

Solution : Let y  = $$e^{2x}$$

$$d\over dx$$ (y) = $$d\over dx$$ $$e^{2x}$$

By using chain rule,

$$d\over dx$$ (y) = 2$$e^{2x}$$

Hence, $$d\over dx$$ ($$e^{2x}$$) = 2$$e^{2x}$$

### (2) Differentiation of $$a^x$$ :

The differentiation of $$a^x$$ with respect to x is $$a^x log_e a$$.

i.e. $$d\over dx$$ $$a^x$$ = $$a^x log_e a$$

#### Proof Using first Principle :

Let f(x) = $$a^x$$. Then, f(x + h) = $$a^{x + h}$$

$$\therefore$$   $$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$f(x + h) – f(x)\over h$$

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$a^{x + h} – a^x\over h$$

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$a^x.a^h – a^x\over h$$

$$d\over dx$$(f(x)) = $$lim_{h\to 0}$$ $$a^x$$ ($$a^h – 1\over h$$)

$$\implies$$ $$d\over dx$$(f(x)) = $$a^x$$  $$lim_{h\to 0}$$ ($$a^h – 1\over h$$)

because, [$$lim_{h\to 0}$$($$a^h – 1\over h$$) = $$log_e a$$]

$$\implies$$ $$d\over dx$$(f(x)) = $$a^x$$ $$\times$$ $$log_e a$$ = $$a^x$$ $$log_e a$$

Hence, $$d\over dx$$ ($$a^x$$) = $$a^x$$ $$log_e a$$

Example : What is the differentiation of $$5^{x}$$ ?

Solution : Let y  = $$5^{x}$$

$$d\over dx$$ (y) = $$d\over dx$$ $$5^{x}$$

$$d\over dx$$ (y) = $$5^x log_e 5$$

Hence, $$d\over dx$$ ($$5^{x}$$) = $$5^x log_e 5$$

### Related Questions

What is the differentiation of $$e^{sinx}$$ ?

What is the integration of $$e^x$$ ?