Jan 26, 2017 · 1/e lnx=-1=>log_(e)x=-1 =>e^(-1)=x :.x=1/e. 140924 views around the world ...

May 30, 2017 · So, we turn #(lnx)^(lnx)# into #e^((ln(lnx))(lnx))#. This works because, due to raising e to any power is an inverse function to taking the natural log of any number, …

Mar 22, 2016 · e^lnx=x let y=e^lnx ln y=lne^lnx->Take ln of both sides lny = lnx * ln e -> use the property log_b x^n ...

Nov 6, 2016 · The derivative of x^(lnx) is [(2*y*(lnx)*(x^(lnx)))/x] let y =x^(lnx) There are no rules that we can apply to easily differentiate this equation, so we just have to mess with it until we …

Jun 30, 2017 · See below. f(x) = (lnx)^2 lnx is defined for x>0 hence, f(x) is defined x>0 lim_(x-> 0) f(x) = +oo and lim_(x->oo) f(x) =+oo f'(x) = 2lnx*(1/x) {Chain rule] For a ...

Feb 6, 2016 · #color(brown)("Total rewrite as changed my mind about pressentation.")# #color(blue)("Preamble:")# Consider the generic case of #" "log_10(a)=b#

Nov 5, 2017 · The question is to find the value of #lnx/x# where #x to oo# If we let #x=oo# then #lnx/x=(oo)/(oo)# #=# Undefined . So now we can apply L'Hospital's rule by differentiating the …

Mar 20, 2018 · Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e. 1 Answer . Rhys

Jul 21, 2015 · How do you find the taylor series series for #f(x)=lnx# at a=2? Calculus Power Series Constructing a Taylor Series. 2 Answers

May 29, 2017 · 1 We can also do this without first using the identity e^lnx=x, although we will have to use this eventually. Note that d/dxe^x=e^x, so when we have a function in the exponent the …