Please correct my proof for the derivative of log x

  • Thread starter galimax
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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.
 

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