2007-01-12 13:35:48 · 8 answers · asked by bklyn 1 in Science & Mathematics ➔ Mathematics
f(x)=log_10 ( x ) = ln x / ln 10
so
f'(x) = 1/(x ln 10) .
2007-01-16 08:34:11 · answer #1 · answered by Anonymous · 4⤊ 0⤋
You must first change the logarithm to base e. To do this, rewrite as 10^y=x. Take ln of both sides and apply log properties to acheive yln10=lnx. Now differentiate implicitly.
dy/dx=1/(xln10)
2007-01-13 00:55:23 · answer #2 · answered by Pius Thicknesse 4 · 2⤊ 0⤋
Ouch, I suggest you look in an advanced maths textbook for something that involves differentiating logs with bases that are not e.
2007-01-12 21:52:58 · answer #3 · answered by samuelll 2 · 2⤊ 0⤋
y = log10x
if it is to the base of 10, then here's how you find the inverse:
- replace x with y and isolate for y
x = log10y
10^x = y
I think that's the right answer..
2007-01-12 22:04:53 · answer #4 · answered by Bored 2 · 1⤊ 1⤋
f(x)= Open you calc book.. It has the answer in the cover . Why are we doing your homework ? I the real world .... not so much .!
2007-01-12 22:48:15 · answer #5 · answered by Titanium_Diboride 2 · 2⤊ 1⤋
f'(x) = 1/(10x) * (10) = 1/x
2007-01-12 21:48:25 · answer #6 · answered by JasonM 7 · 2⤊ 1⤋
f(x)=log₁₀ x = ln x / ln 10
f'(x) = 1/(x ln 10)
2007-01-12 22:26:56 · answer #7 · answered by sahsjing 7 · 2⤊ 0⤋
d(log x) = 1/x dx
Now, use the chain rule and figure it out yourself.
2007-01-12 21:47:34 · answer #8 · answered by Anonymous · 3⤊ 1⤋