Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please correct my proof for the derivative of log x

  1. Sep 3, 2012 #1
    I found this proof on a website for the derivative of ln x:

    y = ln x.
    e^y = x,
    dy/dx * e^y = 1,
    dy/dx * x = 1,
    dy/dx = 1/x.

    My question is, why can't we use a similar method to prove the derivative of log(b)x = 1/x, like this:

    y = log(b) x.
    b^y = x,
    dy/dx * b^y = 1,
    dy/dx * x = 1,
    dy/dx = 1/x.

    Thanks so much everyone.
     
  2. jcsd
  3. Sep 3, 2012 #2
    The differentiation rule only applies to e^x. If you think about it, b^x=(e^lnb)^x = e^(xlnb). Clearly the derivative of this is not the same as b^x.
     
  4. Sep 3, 2012 #3
    Thanks Klungo, I get it now. I did not differentiate the left side in line 2 properly with respect to x.

    Just to document the solution:

    Instead of:
    y = log(b) x.
    b^y = x,
    dy/dx * b^y = 1,
    dy/dx * x = 1,
    dy/dx = 1/x.

    it should be:

    y = log(b) x.
    b^y = x,
    dy/dx * b^y * lnb = 1,
    dy/dx * x * lnb = 1,
    dy/dx = 1/(x * lnb).

    This is because when we differentiate b^y(x) with respect to x, it is as though we differentiate h(x) = h(y(x)). Thus we must use the chain rule, and get:
    h'(x) = h'(y) * dy/dx.
    that is,
    d/dx b^y = b^y * lnb * dy/dx.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Please correct my proof for the derivative of log x
Loading...