Math, asked by anishamohan413, 5 hours ago

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

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Answers

Answered by sakibrez690
1

Answer:

y=a^x

log on both side

logy=loga^x

y=xloga

Answered by mathdude500
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}}}

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