Differentiate the function f(x) = log[2] (1-3x)

Question

ash2326 · Answer

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
$$f'(x)=\frac{3}{\ln 2}\frac{1}{3x-1}$$