2016年4月3日 · Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Calculators. 2 Answers . Shura

2016年1月15日 · According to the chain rule,. #d/dx[lnu]=1/u*u'# Thus, #d/dx[ln2x]=1/(2x)*d/dx[2x]# #=1/(2x)*2=1/x# Another way to think about this problem is to first split up the ...

2009年10月17日 · Homework Statement What is the derivative of ln(2x)? I was just thinking about this, and I got 2 answers. I am in Calc 2 right now. Homework Equations Derivative of ln(x) = 1/x The Attempt at a Solution Since d/dx lna = (1/a)*(derivative of a) Thus d/dx ln2x = (1/2x)*(2)...

2016年4月3日 · How do you find the derivative of #y=(ln2x)^2#? Calculus Basic Differentiation Rules Chain Rule. 1 Answer

2017年9月23日 · 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 =e^2 Hence, ln^2 x = ln x^2 is only true for x=e^2

2015年5月25日 · We can use the chain rule here, naming #u=2x# and remembering that the chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#

2016年11月24日 · x(ln(2x)-1)+C >I=intln(2x)dx We should use integration by parts in the absence of all other possible integration strategies.

2015年3月16日 · You can evaluate this integral by using integration by parts. Let u=ln(2x) Let dv=1/x^2dx Differentiating u=ln(2x) we have du=1/xdx Integrating dv=1/x^2dx we have intdv=intx^-2dx v=-1/x Recall the integration by parts formula uv-intvdu Now proceed as follows -ln(2x)/x-int-1/x(1/x)dx -ln(2x)/x-int-1/x^2dx -ln(2x)/x+intx^-2dx -ln(2x)/x-1/x+C We can rewrite if you like -((ln(2x)+1))/x+C General ...

2014年7月24日 · This is the composite of ln x and 2x, so we use the Chain Rule together with the facts that (2x)'=2 and that (ln x)'=1/x: (ln(2x))'=1/(2x) \times (2x)'=2/(2x)=1/x.

2015年11月30日 · x = 2/e We will use the following: ln(a^x) = xln(a) ln(a) - ln(b) = ln(a/b) e^ln(a) = a 2ln(x) + 1 = ln(2x) => ln(x^2) + 1 = ln(2x) (by the first property above ...