I have a problem with solving derivative of $f(x)$ in this case: $$f(x) = x\times 10^9 - 2^{\log_{10} x\times 10^9}$$ This is what I have: $$f^\prime(x) = \lim_{m\to0} {f(x+m) - f(x)\over m}$$ $$= \lim_{m\to0} {{((x+m)\times 10^9 - 2^{\log_{10} (x+m)\times 10^9}) - (x\times 10^9 - 2^{\log_{10} x\times 10^9})} \over m}$$ $$= \lim_{m\to0} {{m\times10^9 - 2^{\log_{10} (x+m)\times 10^9} + 2^{\log_{10} x\times10^9}} \over m}$$ From here I can't move. I can probably calculate ${m \over m}$ to have problem in this state $$= \lim_{m\to0} 10^9 + {{- 2^{\log_{10} (x+m)\times 10^9} + 2^{\log_{10} x\times10^9}} \over m}$$ But if $m$ becomes $0$, it looks like the result is...
$$f^\prime(x) = 10^9+0$$ ...because ${x \over 0}=0$.
What am I doing wrong? How can I solve this?
Please help.
First you have to see that is a mistake to think that $\frac{x}{0}=0$, you can not say that. If you have a limit process like : $\displaystyle\lim_{x\to 0^+}1/x=\infty$. But let's go to the correct answer :
First note that $f(x)= x\times 10^9 - 2^{\log_{10}x\times 10^9}$ can be separated in two functions, first $h(x)= x\times 10^9$ and $g(x)=2^{\log_{10}x\times 10^9}$, so $f(x)=h(x)-g(x)$ it's clear also that $f'(x)=h'(x)-g'(x)$ and $h'(x)=10^9$ so let's calculate $g'(x)$, we first use a simple property of logarithm that change the product into sum, so $$g(x)=2^{\log_{10}x\times 10^9}= 2^{\log_{10}10^9+\log_{10}x=2^9\times 2^{log_{10}x}}$$ so our work is to derivate $r(x)=2^{log_{10}x}$ now you need to know that $$(a^x)'=a^x\ln a \;\;\;\mbox{and}\;\;\; (\log_{b} x)'= \frac{1}{x\ln b}$$ and now you apply the chain rule as follows:
$$r'(x) = (2^{\log_{10}x})'= 2^{\log_{10}x}\ln 10 \frac{1}{x \ln 10}=\frac{2^{\log_{10}x}}{x}$$
hence, $g'(x)=2^9\times r'(x)=\frac{2^{\log_{10}x\times 10^9}} {x}$
so $f'(x)=h'(x)-g'(x)=10^9-\frac{2^{\log_{10}x\times 10^9}} {x}$