2014年12月15日 · 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 …

2016年11月6日 · 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 …

2017年1月26日 · 1/e lnx=-1=>log_(e)x=-1 =>e^(-1)=x :.x=1/e. 138745 views around the world ...

2007年12月8日 · In summary, deriving the series expansion of ln(x) involves expanding the function about a specific point, typically x = 1. The series expansion formula is …

2015年5月20日 · 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 ...

2015年3月24日 · You know that if #x>1 ln(x)>0# so the limit must be positive. You also know that #ln(x_2)-ln(x_1)=ln(x_2/x_1)# so if #x_2>x_1# the difference is positive, so #ln(x)# is always …

2015年12月19日 · x = 1/e^5 From the definition of a logarithm, we have the property e^ln(x) = x. From this: ln(x) = -5 => e^(ln(x)) = e^(-5) => x = 1/e^5

2015年11月11日 · It is exactly x. You are looking for a number that is the exponent of the base of ln which gives us the integrand, e^x; so: the base of ln is e; the number you need to be the …

2015年11月14日 · Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e. 1 Answer . mason m

2016年7月29日 · #log_e x+log_e(x-3) = log_e 1# so #log_e x(x-3) = log_e 1# or. #x(x-3)=1 -> x^2-3x-1=0# with the solutions. #x = (3 pm sqrt(13))/2# now, cheking the feasibility we choose. #x …