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

$$showthat$$$$iff\left(x\right)=a^x$$$$$$$$showthatf{}^{\prime }\left(x\right)=a^{x}ln\left(a\right)$$$$$$$$byusing\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$$

Commented by niroj last updated on 22/Feb/20

$$\mathrm{Solution}:$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}^{\mathrm{x}}$$$$\mathrm{f}(\mathrm{x}+\mathrm{h})={\mathrm{a}}^{\mathrm{x}+\mathrm{h}}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{h}}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{{\mathrm{a}}^{\mathrm{x}+\mathrm{h}}-{\mathrm{a}}^{\mathrm{x}}}{\mathrm{h}}$$$$=\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{{\mathrm{a}}^{\mathrm{x}}({\mathrm{a}}^{\mathrm{h}}-1)}{\mathrm{h}}$$$$={\mathrm{a}}^{\mathrm{x}}.\underset{\mathrm{h}\to 0}{\stackrel{\lim }{}}\frac{{\mathrm{a}}^{\mathrm{h}}-1}{\mathrm{h}}$$$$={\mathrm{a}}^{\mathrm{x}}\log \mathrm{a}\because (\underset{\mathrm{x}\to 0}{\stackrel{\lim }{}}\frac{{\mathrm{a}}^{\mathrm{x}}-1}{\mathrm{x}}=\log \mathrm{a})$$

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

$$\underset{h\to 0}{\lim }\frac{a^{x+h}-a^x}{h}$$$$a^x\times \underset{h\to 0}{\lim }\frac{a^h-1}{h}[now{a}^h=e^{hlna}]$$$$a^x\times \underset{h\to 0}{\lim }\frac{e^{hlna}-1}{hlna}\times lna$$$$a^x\times 1\times lna$$$$a^{x}lna$$

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