Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 4 years ago

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

  • This Question is Closed
  1. ash2326
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy