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 \(e^x\) ?

Leave a Comment

Your email address will not be published.