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How do you rearrange u equals 1 plus lnx to get x equals?
If: u = 1+lnx Then: x = (u-1)/(ln)
What is the derivative of y equals xln x?
(xlnx)' = lnx + 1
Lnx plus Ln2 equals 2 plus lnx?
NO! Lnx + Ln2= 2 + Lnx implies Ln2 = 2 which implies 2 = e2 which is simply not true.
What is the second derivative of x2 ln 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
What is the derivative of lnx raised to 4?
ln(x4)?d/dx(ln(u))=1/u*d/dx(u)d/dx(ln(x4))=[1/x4]*d/dx(x4)-The derivative of x4 is:d/dx(x4)=4x4-1d/dx(x4)=4x3d/dx(ln(x4))=[1/x4]*(4x3)d/dx(ln(x4))=4x3/x4d/dx(ln(x4))=4/x(lnx)4?Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(lnx)4=4(lnx)4-1*d/dx(lnx)d/dx(lnx)4=4(lnx)3*d/dx(lnx)-The derivative of lnx is:d/dx(ln(u))=1/u*d/dx(u)d/dx(lnx)=1/x*d/dx(x)d/dx(lnx)=1/x*(1)d/dx(lnx)=1/xd/dx(lnx)4=4(lnx)3*(1/x)d/dx(lnx)4=4(lnx)3/x
How do you rearrange u equals 1 plus lnx to get x equals?
If: u = 1+lnx Then: x = (u-1)/(ln)
What is the derivative of y equals xln x?
(xlnx)' = lnx + 1
What is the derivative of lnx raised to lnx?
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
Lnx plus Ln2 equals 2 plus lnx?
NO! Lnx + Ln2= 2 + Lnx implies Ln2 = 2 which implies 2 = e2 which is simply not true.
What is the derivative of 1-lnx divided by 2x?
-1
Simplify e to the 1 plus Inx power?
e1 + (lnx) = e1 * e(lnx) = e * x = ex
What is the derivative of 1 lnx?
The derivative of 1/lnx, can be found easily using either the chain rule or the quotient rule. It is -1/[x*(lnx)2]
What is the second derivative of x2 ln 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
What is the derivative of lnx raised to 4?
ln(x4)?d/dx(ln(u))=1/u*d/dx(u)d/dx(ln(x4))=[1/x4]*d/dx(x4)-The derivative of x4 is:d/dx(x4)=4x4-1d/dx(x4)=4x3d/dx(ln(x4))=[1/x4]*(4x3)d/dx(ln(x4))=4x3/x4d/dx(ln(x4))=4/x(lnx)4?Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(lnx)4=4(lnx)4-1*d/dx(lnx)d/dx(lnx)4=4(lnx)3*d/dx(lnx)-The derivative of lnx is:d/dx(ln(u))=1/u*d/dx(u)d/dx(lnx)=1/x*d/dx(x)d/dx(lnx)=1/x*(1)d/dx(lnx)=1/xd/dx(lnx)4=4(lnx)3*(1/x)d/dx(lnx)4=4(lnx)3/x
Find all solutions of equation lnx plus lnx plus 3 equals ln55divided by 4?
lnx + lnx +3 = (ln55)/4 2lnx +3 =(ln55)/4 8lnx + 12=ln55 8lnx=-12+ln55 lnx=(-12+ln55)/8 x=e^[(-12+ln55)/8]
What is the Derivative of x to the power of x?
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
What is the integral of x.lnx?
Do you want ∫x lnx dx? Let's call that I, which we now seek to find. The solution is I = ½ x2 lnx - ¼ x2; here is how we can find it: Let z = lnx. Then, x = ez, dx = ez dz, and dI = x lnx dx. Then, dI = (ez)(z)(ez dz) = ze2z dz = ¼ (2z e2z d(2z)). Thus, 4dI = wew dw, where we let w = 2z = 2 lnx. Now, d(wew) = ew dw + w d(ew) = ew dw + wew dw = ew dw + 4dI; hence, 4dI = d(wew) - ewdw = d(wew) - d(ew)­ = d((w - 1)ew). Therefore, 4I = (w - 1)ew = (2 lnx - 1)x2 = 2x2 lnx - x2; and I = ½ x2 lnx - ¼ x2, which is the answer we sought. Checking, we differentiate back, to confirm that I' = x lnx: I = ¼ x2(2 lnx - 1), whence, 4I' = (x2(2 lnx - 1))' = 2x(2 lnx - 1) + x2 (2/x) = 2x(2 lnx - 1) + 2x = 2x(2 lnx) = 4x lnx; thus, I' = x lnx, re-assuring us that we have integrated correctly.
