I thought it would be interesting to derive d/(dx) (ln x)
We know the definition of the derivative is:
(d(f(x)))/dx = lim_(h to 0) (f(x+h)-f(x))/h
=> d/(dx) ( lnx) = (ln(x+h) - lnx )/h
Using our log laws:
lim_(h to 0) (ln( (x+h)/x ) )/h
We know (x+h)/x = 1 + h/x
=> lim_(h to 0) (ln(1+h/x))/h
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The not so interesting way:
As h to 0 the denominator and numirator -> 0
So know we can use the L'Hopitals rule:
=> lim_(h to 0 ) (d/(dh) ( ln(1+h/x) ))/( d/(dh)( h ))
=> lim_(h to 0 ) ((1/x)/(1+h/x) )/1
= 1/x
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Or we could use the initial way:
=> lim_(h to 0) (1+h/x)^(1/h) = e^(1/x)
As we know:
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lim_(phi to oo) (1+ 1/(phi))^phi = e
lim_(phi to 0) (1+ phi)^(1/phi) = e
therefore lim_(phi to 0) (1+ phi/gamma )^(gamma/phi) = e
Raising each side to 1/gamma power
lim_(phi to 0) (1 + phi/gamma)^(1/phi) = e^(1/gamma)
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Raising each side by the power of h:
=> lim_(h to 0) (1+h/x) = e^(h/x)
=> lim_(h to 0) ln(e^(h/x)) /h
=> lim_(h to 0) ((h/x)* ln(e))/h
( ln(e) = 1 )
=> lim_(h to 0) 1/x
= 1/x
