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Derivative of lnx?
1/X
What is the derivative of y equals xln x?
(xlnx)' = lnx + 1
What is the derivative and second derivative of lnx to the third power?
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
What is the derivative of ln 1 plus x?
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
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
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]
Derivative of lnx?
1/X
What is the Second derivative of lnx?
-1/x2
What is the derivative of 5x?
The derivative of 5x is 5.
What is the derivative of 1-lnx divided by 2x?
-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
What is the derivative of lnx squared?
I do not see why the chain rule would not work here. d/dx (inx)^2 = 2(lnx) * 1/x = 2(lnx)/x
What is the derivative of natural log?
d/dx lnx=1/x
What is the anti derivative of lnx?
x (ln x + 1) + Constant
What is the derivative of y equals xln x?
(xlnx)' = lnx + 1
Integral of xlnxdx?
integration by parts. Let u=lnx, dv=xdx-->du=(1/x)dx, v=.5x^2. Integral of (xlnxdx)=lnx*.5x^2-integral of .5x^2(1/x)dx=lnx*.5x^2-integral of .5xdx=lnx*.5x^2-(1/6)x^3. That's it.
What is the derivative and second derivative of lnx to the third power?
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
