Find the derivative of a^x w.r.t.x (a>0) using first principle (from the definition). About the author Adalynn
[tex]\large\underline{\sf{Solution-}}[/tex] [tex]\rm :\longmapsto\:Let \: f(x) = {a}^{x} [/tex] So, [tex]\rm :\longmapsto\ \: f(x + h) = {a}^{x + h} [/tex] Now, By using definition, we have [tex]\rm :\longmapsto\:f'(x) = \lim_{h \to 0}\dfrac{ {a}^{x + h} – {a}^{x} }{h} [/tex] [tex] \rm \: \: = \: \lim_{h \to 0}\dfrac{ {a}^{x} \times {a}^{h} – {a}^{x} }{h} [/tex] [tex] \rm \: \: = \: \lim_{h \to 0}\dfrac{ {a}^{x}({a}^{h} – 1)}{h} [/tex] [tex] \rm \: \: = \: {a}^{x} \lim_{h \to 0}\dfrac{ {a}^{h} – 1 }{h} [/tex] [tex] \rm \: \: = \: {a}^{x} \times log(a) [/tex] [tex] \: \: \: \: \red{\bigg \{ \because \: \lim_{x \to 0}\dfrac{ {a}^{x} – 1}{x} = log(a) \bigg \}}[/tex] [tex] \rm \: \: = \: {a}^{x} log(a) [/tex] Hence, [tex]\bf\implies \:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) [/tex] Additional Information :- [tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{sinx}{x} = 1}}}[/tex] [tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{tanx}{x} = 1}}}[/tex] [tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {e}^{x} – 1}{x} = 1}}}[/tex] [tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {a}^{x} – 1}{x} = log(a) }}}[/tex] [tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{log(1 + x)}{x} = 1}}}[/tex] [tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {sin}^{ – 1} x}{x} = 1}}}[/tex] [tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {tan}^{ – 1} x}{x} = 1}}}[/tex] [tex]\green{\boxed{ \bf{ \:\lim_{x \to 0} \: x \: sin\dfrac{1}{x} = 0}}}[/tex] Reply
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex]\rm :\longmapsto\:Let \: f(x) = {a}^{x} [/tex]
So,
[tex]\rm :\longmapsto\ \: f(x + h) = {a}^{x + h} [/tex]
Now,
By using definition, we have
[tex]\rm :\longmapsto\:f'(x) = \lim_{h \to 0}\dfrac{ {a}^{x + h} – {a}^{x} }{h} [/tex]
[tex] \rm \: \: = \: \lim_{h \to 0}\dfrac{ {a}^{x} \times {a}^{h} – {a}^{x} }{h} [/tex]
[tex] \rm \: \: = \: \lim_{h \to 0}\dfrac{ {a}^{x}({a}^{h} – 1)}{h} [/tex]
[tex] \rm \: \: = \: {a}^{x} \lim_{h \to 0}\dfrac{ {a}^{h} – 1 }{h} [/tex]
[tex] \rm \: \: = \: {a}^{x} \times log(a) [/tex]
[tex] \: \: \: \: \red{\bigg \{ \because \: \lim_{x \to 0}\dfrac{ {a}^{x} – 1}{x} = log(a) \bigg \}}[/tex]
[tex] \rm \: \: = \: {a}^{x} log(a) [/tex]
Hence,
[tex]\bf\implies \:\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) [/tex]
Additional Information :-
[tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{sinx}{x} = 1}}}[/tex]
[tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{tanx}{x} = 1}}}[/tex]
[tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {e}^{x} – 1}{x} = 1}}}[/tex]
[tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {a}^{x} – 1}{x} = log(a) }}}[/tex]
[tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{log(1 + x)}{x} = 1}}}[/tex]
[tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {sin}^{ – 1} x}{x} = 1}}}[/tex]
[tex]\green{\boxed{ \bf{ \:\lim_{x \to 0 \: }\dfrac{ {tan}^{ – 1} x}{x} = 1}}}[/tex]
[tex]\green{\boxed{ \bf{ \:\lim_{x \to 0} \: x \: sin\dfrac{1}{x} = 0}}}[/tex]