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: # …

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^ln (sin (2x))= e^0; 0 < x < pi/2# The exponential function and the natural logarithm are inverses, therefore, they cancel and any number to the 0 power is 1:

Please see the explanation. Let y = sqrt (x)^x Because the square root is the same as the 1/2 power, we can write the above equation as: y = (x^ (1/2))^x A property of exponents allows us to multiply them: …

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))= …

Jan 16, 2018 · dy/dx = (2x^lnxlnx)/x I'm pretty fond of logarithmic differentiation. We try to take the derivative of both sides: lny = ln(x^(lnx)) lny = lnxlnx Now use implicit differentiation and the product …

You can do that by writing the Arrhenius equation in non-exponential form ln (k) = ln [A * "exp" (-E_a/ (RT))] This is equivalent to ln (k) = ln (A) + ln ["exp" (-E_a/ (RT))] ln (k) = ln (A) - E_a/ (RT) Now, …

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 ...

LEFT-HAND INTEGRAL The left-hand integral is trivial, as there are a lot of constants: (epsilonA_ (s,r)sigma)/ (rhoVc)int_ (0)^ (t)dt = (epsilonA_ (s,r)sigma)/ (rhoVc ...

(2) So, the Taylor series for ln (x) centered at x=1 is given by ln (x) = sum_ {n=0}^oo f^ ( (n)) (1)/ (n!) (x-1)^n qquadqquad = sum_ {n=0}^oo (-1)^ (n-1) ( (n-1)!)/ (n!) (x-1)^n qquad qquad = sum_ {n=0}^oo ( …