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, #e^ln(x)=x# for any x. This may seem like a more complicated function we have made to take the derivative, but trust me, it makes thins simpler.

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 find an answer. If we take the natural log of both sides, we are changing the equation. We can do this as long as we take into account that this will be a completely new equation: lny=ln(x^(lnx)) …

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 numerator and denominator individually. So, #d/dxlnx=1/x# and #d/dxx=1# #lim_(x -> oo)(1/x)/1# #lim_(x -> oo)1/x# Now we let #x=oo#

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 chain rule will apply: d/dxe^u=e^u*(du)/dx. So: d/dxe^lnx=e^lnx(d/dxlnx) The derivative of lnx is 1/x: d/dxe^lnx=e^lnx(1/x) Then using the identity e^lnx=x: d/dxe^lnx=x(1/x)=1 Which is the …