Here's the question you clicked on:
calyne
Differentiate the function f(x) = log[2] (1-3x)
We have \[f(x)=\log_{2} (1-3x)\] notice that the base here is 2, we can't directly differentiate it. We need to have the base of log as e. We know the logarithm's property \[\log_{a} b=\frac{\log_c{b}}{\log_{c} a}\] so here we have \[f(x)= \frac{\log_e(1-3x)}{\log_e {2}}\] Now we can write as \[f(x)=\frac{\ln (1-3x)}{\ln 2}\] Let's differentiate this with respect to x \[f'(x)=\frac{d}{dx}(\frac{\ln (1-3x)}{\ln 2})\] We get \[f'(x)=\frac{1}{\ln 2}\frac{d}{dx}{\ln (1-3x)}=\frac{1}{2} \frac{1}{1-3x} \frac{d}{dx} (1-3x)\] We get \[f'(x)=\frac{1}{\ln 2}\frac{1}{1-3x} \times {-3}\] We get finally \[f'(x)=\frac{3}{\ln 2}\frac{1}{3x-1}\]