Jan 26, 2017 · 1/e lnx=-1=>log_(e)x=-1 =>e^(-1)=x :.x=1/e. 141132 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 ...

Sep 23, 2017 · Yes, but also see below ln^2 x is simply another way of writing (lnx)^2 and so they are equivalent. However, these should not be confused with ln x^2 which is equal to 2lnx There is only one condition where ln^2 x = ln x^2 set out below. ln^2 x = ln x^2 -> (lnx)^2 = 2lnx :. lnx * lnx = 2lnx Since lnx !=0 lnx * cancel lnx = 2 * cancel lnx lnx = 2 x …

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#

Mar 20, 2018 · (dy)/(dx) = 1/(xlnx) d/dx ln f(x) = ( f'(x) ) / f(x) => d/dx( ln ( ln x ) ) = (d/dx( lnx )) /lnx = (1/x)/lnx 1/( xlnx )

Dec 15, 2014 · Lets start by breaking down the function. (ln(x))/x = 1/x ln(x) So we have the two functions; f(x) = 1/x g(x) = ln(x) But the derivative of ln(x) is 1/x, so f(x) = g'(x). This means we can use substitution to solve the original equation. Let u = ln(x). (du)/(dx) = 1/x du = 1/x dx Now we can make some substitutions to the original integral. int ln(x) (1/x dx) = …

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 ...

Mar 4, 2016 · dy/dx = 2/x Applying the chain rule, along with the derivatives d/dx ln(x) = 1/x and d/dx x^2 = 2x, we have dy/dx = d/dxln(x^2) =1/x^2(d/dxx^2) =1/x^2(2x) =2/x

May 20, 2015 · firstly we look at the formula for the Taylor series, which is: f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n which equals: f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/(2!) + (f ...