Jun 11, 2017 · The left hand side of the equation is just the product rule for derivatives.

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

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#

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 …

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 …

Jun 29, 2017 · Write the sum as: bn = 2 2n ∑ j=1 1 2n ln(1 + (2j)2 (2n)2) Consider now the function: f (x) = ln(1 + 4x2) over the interval x ∈ (0,1). As: f '(x) = 2x 1 +x2> 0 in the interval, the …

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 …

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: …

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

Let us find y'. By repeatedly applying Chain Rule, y'=2 [ln (1+e^x)]^1cdot [ln (1+e^x)]' =2ln (1+e^x)cdot {1}/ {1+e^x}cdot (1+e^x)' =2ln (1+e^x)cdot {1}/ {1+e^x}cdot ...