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