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Let $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})$ and $\mathrm{f}(\mathrm{x})=1+\mathrm{xg}(\mathrm{x}) \phi(\mathrm{x})$,
where $\lim _{x \rightarrow 0} g(x)=a$ and $\lim _{x \rightarrow 0} \phi(x)=b$. Then what is $f^{\prime}(x)$
equal to?
Options:
where $\lim _{x \rightarrow 0} g(x)=a$ and $\lim _{x \rightarrow 0} \phi(x)=b$. Then what is $f^{\prime}(x)$
equal to?
Solution:
2534 Upvotes
Verified Answer
The correct answer is:
$\operatorname{abf}(\mathrm{x})$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{f(x) f(h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(x)[f(h)-1]}{h}=\lim _{h \rightarrow 0} \frac{f(x)[h g(h) \phi(h)]}{h}$
$=a b f(x)$
$=\lim _{h \rightarrow 0} \frac{f(x)[f(h)-1]}{h}=\lim _{h \rightarrow 0} \frac{f(x)[h g(h) \phi(h)]}{h}$
$=a b f(x)$
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