What is the derivative of #2ln(x)#?

Question

Ezra Adams · Accepted Answer

The derivative of #ln(x)# is #1/x#. Thus, keeping the constant out of the derivation (it being only a coefficient)... #(dy)/(dx)=2(1/x)=2/x#

Emma Aldridge · Answer

# d/(dx)( lnx) = 1/x #

Using first principles

An expansion on the prior answer

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 # #color(red)(--------------)# 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 # #color(red)(--------------)# Or we could use the initial way: #=> lim_(h to 0) (1+h/x)^(1/h) = e^(1/x)# As we know: #color(blue)(--------------)# #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) # #color(blue)(--------------)# 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 #

Christopher Agresta · Answer

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