If I differentiate log x, do I get 1/x ? Watch

Jack93o
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I was told so, but I don't quite get why.

because the differential of ln x = 1/x

and obviously log x is not the same as ln x
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Mr M
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(Original post by Jack93o)
I was told so, but I don't quite get why.

because the differential of ln x = 1/x

and obviously log x is not the same as ln x
It depends what base you are using. If that is log to base e (it is quite common for log to mean this rather than base 10 and so on) then yes. If it isn't then no.
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Mr M
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y=\log_a x

x=a^y

\ln x = y \ln a

\displaystyle y = \frac{1}{\ln a}{\ln x}

\displaystyle \frac{dy}{dx}=\frac{1}{x \ln a}
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Hasufel
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one way to see this is by putting: y=\ln(x)=log_{e}(x) then x=e^{y}

then, take dx/dy - not dy/dx: \displaystyle \frac{dx}{dy}=e^{y}=e^{\ln(x)}=x

so then, taking the reciprocal: \displaystyle \frac{dy}{dx}= \frac{1}{x}

(not very rigorous, but hopefully easy to follow)
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Kvothe the Arcane
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Another proof in Spoiler
Spoiler:
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\displaystyle \dfrac{d}{dx}lnx=\lim_{h \to 0} \left(\dfrac{ln(x+h)-ln(x)}{h} \right)

\displaystyle  =\lim_{h \to 0} \left(\dfrac{ln \left( \dfrac{x+h}{x} \right)}{h} \right)=\lim_{h \to 0} \left( \dfrac{1}{h} ln \left(1+\dfrac{h}{x} \right) \right)=\lim_{h \to 0} \left( ln \left(1+\dfrac{h}{x} \right)^\frac{1}{h} \right)
sub

let u=\dfrac{h}{x}

After sub, derivative is:
\displaystyle \lim_{u \to 0} \left( ln(1+u)^\frac{1}{u} \right)^\frac{1}{x}
\displaystyle \dfrac{1}{x}\lim_{u \to 0} \left( ln(1+u)^\frac{1}{u} \right)
sub

let n=\dfrac{1}{u}

Derivative becomes
\displaystyle \lim_{n \to \infty} \dfrac{1}{x} \left( ln \left(1+\dfrac{1}{n} \right)^n \right)=\dfrac{1}{x}ln(e)=\dfrac  {1}{x}
\therefore \dfrac{d}{dx}lnx=\dfrac{1}{x}.
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EmilyJones123
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Sorry to interrupt! Please can somebody help!!??? I am probably being really dense but how do you solve 3sinx-4cosx=3 in the range where x is between 0 and 2π??? Thank you ever so much!!
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Mr M
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(Original post by EmilyJones123)
Sorry to interrupt! Please can somebody help!!??? I am probably being really dense but how do you solve 3sinx-4cosx=3 in the range where x is between 0 and 2π??? Thank you ever so much!!
Please don't hijack other threads.

Express 3sin x - 4 cos x in harmonic form (in the form R \sin (x \pm \alpha) or R \cos (x \pm \alpha)).
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