1/X
Oh, dude, the third derivative of ln(x) is -2/(x^3). But like, who really needs to know that, right? I mean, unless you're planning on impressing your calculus teacher or something. Just remember, math is like a puzzle, except no one actually wants to put it together.
First one: f'(x) = 2x*lnx + x^2*(1/x) = 2x*lnx + x = x*(2*lnx + 1) Second derivate: f"(x) = D [x*(2*lnx + 1)] = 1*(2*lnx + 1) + x*(2/x) = 2*lnx+1+2 = 2*lnx + 3 So, there is a minimum in this graph on point (1/e^(1/2)), -1/(2e)) = MIN(x, y) Van Sanden David Belgium
(xlnx)' = lnx + 1
There are 2 interpretations of your question: First: e^[lnx + lny] =e^[ln(xy)] =xy Second: lny + e^(lnx) =lny + x
-1/x2
The solution to this is: (xx)'= (elnx to the power of x)'= (exlnx)'= (xlnx)'*exlnx= [x(1/x) + 1(lnx)]*exlnx = (lnx+1)*exlnx= (lnx+1)*xx
The derivative of 1/lnx, can be found easily using either the chain rule or the quotient rule. It is -1/[x*(lnx)2]
1/X
Oh, dude, the third derivative of ln(x) is -2/(x^3). But like, who really needs to know that, right? I mean, unless you're planning on impressing your calculus teacher or something. Just remember, math is like a puzzle, except no one actually wants to put it together.
-1
start by setting y=lnx^lnx take ln of both sides lny=lnx(ln(lnx)) differentiate dy/dx(1/y)=(1+ln(lnx))/x dy/dx=y(1+ln(lnx))/x we know that y=lnx^lnx so we can just substatute back in dy/dx=(lnx^lnx)*(1+ln(lnx))/x
I get x*x^x-1 + lnx*x^x = x^x + x^xlnx = x^x * (1+lnx) Here, ^ is power; * = times; ln = natural logratithm ( base e)
I do not see why the chain rule would not work here. d/dx (inx)^2 = 2(lnx) * 1/x = 2(lnx)/x
First one: f'(x) = 2x*lnx + x^2*(1/x) = 2x*lnx + x = x*(2*lnx + 1) Second derivate: f"(x) = D [x*(2*lnx + 1)] = 1*(2*lnx + 1) + x*(2/x) = 2*lnx+1+2 = 2*lnx + 3 So, there is a minimum in this graph on point (1/e^(1/2)), -1/(2e)) = MIN(x, y) Van Sanden David Belgium
d/dx lnx=1/x
x (ln x + 1) + Constant