If you mean: y =(lnx)3
then:
dy/dx = [3(lnx)2]/x
ddy/dx = [(6lnx / x) - 3(lnx)2] / x2
If you mean: y = ln(x3)
Then:
dy/dx = 3x2/x3 = 3/x = 3x-1
ddy/dx = -3x-2 = -3/x2
The first derivative of ln x is 1/x, which (for the following) you better write as x-1.Now use the power rule:Second derivative (the derivative of the first derivative) is -1x-2, the third derivative is the derivative of this, or 2x-3. You may now wish to write this in the alternative form, as 2 / x3.
1/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
(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 first derivative of ln x is 1/x, which (for the following) you better write as x-1.Now use the power rule:Second derivative (the derivative of the first derivative) is -1x-2, the third derivative is the derivative of this, or 2x-3. You may now wish to write this in the alternative form, as 2 / x3.
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
-1
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)
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 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