Q: What is the Second derivative of lnx?

Write your answer...

Submit

Still have questions?

Continue Learning about Other Math

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

Derivative of lnx= (1/x)*(derivative of x) example: Find derivative of ln2x d(ln2x)/dx = (1/2x)*d(2x)/dx = (1/2x)*2===>1/x When the problem is like ln2x^2 or ln-square root of x...., the answer won't come out in form of 1/x.

All it means to take the second derivative is to take the derivative of a function twice. For example, say you start with the function y=x2+2x The first derivative would be 2x+2 But when you take the derivative the first derivative you get the second derivative which would be 2

The formula for finding the derivative of a log function of any "a" base is (dy/dx)log base a (x) = 1/((x)ln(a)) If we're talking about base "e" (natural logs) the answer is 1/(x-2) I think you're asking for the derivative of y = logx2. It's (-logx2)/(x(lnx)).

start by differentiating each component d/dx [x2]=2x d/dx [lnx]=1/x product rule 2xlnx+x2/x simplify (factoring) x[2ln(x)+1]

Related questions

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

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

-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 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

(xlnx)' = lnx + 1

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

f(x)=xln(x) this function is treated as u*v u=x v=ln(x) The derivative of a product is f'(x)=u*v'+v*u' plugging the values back in you get: f'(x)=(x*dlnx/x)+(ln*dx/dx) The derivative of lnx=1/x x=u dlnu/dx=(1/u)*(du/dx) dx/dx=1 x=u dun/dx=nun-1 dx1/dx=1*x1-1 = x0=1 f'(x)=x*(1/x)+lnx*1 f'(x)=1+lnx Now for the second derivative f''(x)=d1/dx+dlnx/dx the derivative of a constant, such as 1, is 0 and knowing that the derivative of lnx=1/x you get f''(x)=(1/x)