Explanation: This integral may seem tricky at first, but it comes apart quite quickly once you realize a certain u-substitution. If we let #u=ln (ln (x))#, we get by the chain rule that the derivative is: # …

Taking logs of both sides, # lne^x=ln [1/sqrt [e-1]]# #but ln [e^x]=x# and so #x=ln [1/sqrt [e-1]]#. Hope this helps. Answer link

Rule 1: sqrt {"nonnegative"} 7-ln (x-1) ge0 by adding ln (x-1), => 7 ge ln (x-1) by raising e to both sides, => e^7 ge e^ {ln (x-1)}=x-1 by adding 1, => e^7+1 ge x Rule2: ln ("positive") x-1 > 0 by adding 1, => x …

E (n) = (sum_ (i=1)^n 1/i) / n E (5) = 0.45dot (6) E (100) ~~ 0.05187 E (n) = (sum_ (i=1)^n 1/i) / n = (H (n))/n where H (n) is the nth Harmonic number. As n -> oo, H (n) - ln (n) -> gamma, where gamma is …

Differentiate both sides: # (d (ln (y)))/dx = (d (sqrt (cos (x)))ln (sin (x)))/dx# Use the chain rule on the left: #1/ydy/dx = (d (sqrt (cos (x)))ln (sin (x)))/dx# Use the product rule on the right: #1/ydy/dx = (d (sqrt …

int\ -ln (x)\ dx=x-xln (x)+C First, I'll move the minus sign out the front: -int\ ln (x)\ dx When you have a relatively simple integral that you can't rewrite in any way, the classic trick is to use integration by …

What is the nth term of the sequence # ln (2/1),ln (3/2),ln (4/3),.. #? What is the limit as #n rarr oo# Calculus

f^ (') (x) = (cos (x)ln (x)+sin (x)/x)/2 The function given is: =>f (x) = ln (sqrt (x^ (sin (x)))) Let's first simplify sqrt (x^sin (x)). Using the law: u (x) = e^ln (u (x)) " and " ln (a^b) =bln (a) we get: =>sqrt (x^sin (x))= …

1 Answer Guilherme N. Dec 30, 2015 To differentiate the #ln#, we'll need quotient rule. To differentiate #sin (x^2)#, we'll need chain rule as well. To differentiate #sin (x^2)/x#, we'll need quotient rule.

Jan 16, 2018 · Explanation: I'm pretty fond of logarithmic differentiation. We try to take the derivative of both sides: #lny = ln (x^ (lnx))# #lny = lnxlnx#