Question

Find the derivative of a^x w.r.t.x (a>0) using first principle (from the definition).​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: f(x) = {a}^{x}$

So,

$\rm :\longmapsto\ \: f(x + h) = {a}^{x + h}$

Now,

By using definition, we have

$\rm :\longmapsto\:f'(x) = \lim_{h \to 0}\dfrac{ {a}^{x + h} - {a}^{x} }{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{ {a}^{x} \times {a}^{h} - {a}^{x} }{h}$

$\rm \: \: = \: \lim_{h \to 0}\dfrac{ {a}^{x}({a}^{h} - 1)}{h}$

$\rm \: \: = \: {a}^{x} \lim_{h \to 0}\dfrac{ {a}^{h} - 1 }{h}$

$\rm \: \: = \: {a}^{x} \times log(a)$

$\: \: \: \: \red{\bigg \{ \because \: \lim_{x \to 0}\dfrac{ {a}^{x} - 1}{x} = log(a) \bigg \}}$

$\rm \: \: = \: {a}^{x} log(a)$

Hence,

$\bf\implies \:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a)$

Additional Information :-

$\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{sinx}{x} = 1}}}$

$\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{tanx}{x} = 1}}}$

$\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {e}^{x} - 1}{x} = 1}}}$

$\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {a}^{x} - 1}{x} = log(a) }}}$

$\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{log(1 + x)}{x} = 1}}}$

$\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {sin}^{ - 1} x}{x} = 1}}}$

$\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {tan}^{ - 1} x}{x} = 1}}}$

$\green{\boxed{ \bf{ \:\lim_{x \to 0} \: x \: sin\dfrac{1}{x} = 0}}}$