Question

plz help me 11th pcbm maths
find the derivative of a^x w. r. t x using first principle​ plz see the attachment

Answer 1

Answer:

y=a^x

log on both side

logy=loga^x

y=xloga

Answer 2

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

Let assume that

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

Change x ⟼ x + h, we get

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

Using definition of First Principle, we have

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

On substituting the values, we get

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

can be rewritten as

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

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{ {a}^{x}( {a}^{h} - 1)}{h}$

$\rm :\longmapsto\:f'(x) = {a}^{x} \displaystyle\lim_{h \to 0} \frac{{a}^{h} - 1}{h}$

We know that,

$\red{ \boxed{ \rm{ \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga}}}$

So, using this result, we get

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

Hence,

$\red{ \boxed{ \bf{ \: \frac{d}{dx} {a}^{x} = {a}^{x} \: loga}}}$

Additional Information :-

Let solve one more problem of same type!!

Differentiate w. r. t. x using first principle

$\rm :\longmapsto\:f(x) = {a}^{2x + 3}$

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

Given that

$\rm :\longmapsto\:f(x) = {a}^{2x + 3}$

Change x ⟼ x + h, we get

$\rm :\longmapsto\:f(x + h) = {a}^{2x + 2h + 3}$

Using Definition of First Principle, we get

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

On substituting the values, we get

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{ {a}^{2x + 2h + 3} - {a}^{2x + 3} }{h}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{ {a}^{2x + 3}. {a}^{2h} - {a}^{2x + 3} }{h}$

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{ {a}^{2x + 3}( {a}^{2h} - 1)}{h}$

$\rm :\longmapsto\:f'(x) = {a}^{2x + 3} \displaystyle\lim_{h \to 0} \frac{{a}^{2h} - 1}{h}$

$\rm :\longmapsto\:f'(x) = {a}^{2x + 3} \displaystyle\lim_{h \to 0} \frac{{a}^{2h} - 1}{2h} \times 2$

$\rm :\longmapsto\:f'(x) = 2{a}^{2x + 3} \displaystyle\lim_{h \to 0} \frac{{a}^{2h} - 1}{2h}$

We know that

$\red{ \boxed{ \rm{ \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga}}}$

$\rm :\longmapsto\:f'(x) = 2{a}^{2x + 3} \: loga$

Hence,

$\red{ \boxed{ \bf{ \: \frac{d}{dx} {a}^{2x + 3} = 2 \: {a}^{2x + 3} \: loga}}}$