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Question Number 82546 by M±th+et£s last updated on 22/Feb/20

s h o w t h a t

i f f (x) = a^{x}

s h o w t h a t f^{'} (x) = a^{x} l n (a)

b y u s i n g \underset{h \to 0}{l i m} \frac{f (x + h) - f (x)}{h}

Commented by niroj last updated on 22/Feb/20

Solution :

f (x) = a^{x}

f (x + h) = a^{x + h}

f^{^{'}} (x) = \underset{h \to 0}{\overset{\lim}{}} \frac{f (x + h) - f (x)}{h}

f^{^{'}} (x) = \underset{h \to 0}{\overset{\lim}{}} \frac{a^{x + h} - a^{x}}{h}

= \underset{h \to 0}{\overset{\lim}{}} \frac{a^{x} (a^{h} - 1)}{h}

= a^{x} . \underset{h \to 0}{\overset{\lim}{}} \frac{a^{h} - 1}{h}

= a^{x} \log a ∵ (\underset{x \to 0}{\overset{\lim}{}} \frac{a^{x} - 1}{x} = \log a)

Answered by TANMAY PANACEA last updated on 22/Feb/20

\lim_{h \to 0} \frac{a^{x + h} - a^{x}}{h}

a^{x} \times \lim_{h \to 0} \frac{a^{h} - 1}{h} [n o w a^{h} = e^{h l n a}]

a^{x} \times \lim_{h \to 0} \frac{e^{h l n a} - 1}{h l n a} \times l n a

a^{x} \times 1 \times l n a

a^{x} l n a