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

show that if f(x)=a^x show that f ′(x)=a^x ln(a) by using lim_(h→0) ((f(x+h)−f(x))/h)

$${show}\:{that}\: \\ $$$${if}\:{f}\left({x}\right)={a}^{{x}} \\ $$$$ \\ $$$${show}\:{that}\:{f}\:'\left({x}\right)={a}^{{x}} \:{ln}\left({a}\right) \\ $$$$ \\ $$$${by}\:{using}\:\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$

Commented by niroj last updated on 22/Feb/20

Solution: f(x)=a^x f(x+h)=a^(x+h) f^( ′ ) (x)= _(h→0) ^(lim) ((f(x+h)−f(x))/h) f^( ′) (x)= _(h→0) ^(lim) ((a^(x+h) −a^x )/h) = _(h→0) ^(lim) ((a^x (a^h −1))/h) = a^x . _(h→0) ^(lim) ((a^h −1)/h) = a^x log a ∵ ( _(x→0) ^(lim) ((a^x −1)/x)=log a)

$$\:\:\:\mathrm{Solution}: \\ $$$$\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}^{\mathrm{x}} \\ $$$$\:\:\:\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)=\mathrm{a}^{\mathrm{x}+\mathrm{h}} \\ $$$$\:\:\:\mathrm{f}^{\:'\:} \left(\mathrm{x}\right)=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\overset{\mathrm{lim}} {\:}}\frac{\mathrm{f}\left(\mathrm{x}+\mathrm{h}\right)−\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}} \\ $$$$\:\:\:\mathrm{f}^{\:'} \left(\mathrm{x}\right)=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\overset{\mathrm{lim}} {\:}}\:\frac{\mathrm{a}^{\mathrm{x}+\mathrm{h}} −\mathrm{a}^{\mathrm{x}} }{\mathrm{h}} \\ $$$$\:\:\:=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\overset{\mathrm{lim}} {\:}}\:\:\frac{\mathrm{a}^{\mathrm{x}} \left(\mathrm{a}^{\mathrm{h}} −\mathrm{1}\right)}{\mathrm{h}} \\ $$$$=\:\:\mathrm{a}^{\mathrm{x}} .\:\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\overset{\mathrm{lim}} {\:}}\:\frac{\mathrm{a}^{\mathrm{h}} −\mathrm{1}}{\mathrm{h}}\:\: \\ $$$$=\:\:\:\mathrm{a}^{\mathrm{x}} \:\mathrm{log}\:\mathrm{a}\:\:\:\:\:\because\:\:\:\left(\underset{\mathrm{x}\rightarrow\mathrm{0}} {\overset{\mathrm{lim}} {\:}}\:\frac{\mathrm{a}^{\mathrm{x}} −\mathrm{1}}{\mathrm{x}}=\mathrm{log}\:\mathrm{a}\right) \\ $$

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

lim_(h→0) ((a^(x+h) −a^x )/h) a^x ×lim_(h→0) ((a^h −1)/h) [now a^h =e^(hlna) ] a^x ×lim_(h→0) ((e^(hlna) −1)/(hlna))×lna a^x ×1×lna a^x lna

$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}^{{x}+{h}} −{a}^{{x}} }{{h}} \\ $$$${a}^{{x}} ×\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}^{{h}} −\mathrm{1}}{{h}}\:\:\left[{now}\:{a}^{{h}} ={e}^{{hlna}} \right] \\ $$$${a}^{{x}} ×\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{hlna}} −\mathrm{1}}{{hlna}}×{lna} \\ $$$${a}^{{x}} ×\mathrm{1}×{lna} \\ $$$${a}^{{x}} {lna} \\ $$

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