0
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)=4x4-1
d/dx(x4)=4x3
d/dx(ln(x4))=[1/x4]*(4x3)
d/dx(ln(x4))=4x3/x4
d/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/x
d/dx(lnx)4=4(lnx)3*(1/x)
d/dx(lnx)4=4(lnx)3/x
What else can I help you with?
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 4th derivative of lnx?
The fourth derivative of ( \ln(x) ) can be determined by first calculating its derivatives. The first derivative is ( \frac{1}{x} ), the second derivative is ( -\frac{1}{x^2} ), the third derivative is ( \frac{2}{x^3} ), and the fourth derivative is ( -\frac{6}{x^4} ). Thus, the fourth derivative of ( \ln(x) ) is ( -\frac{6}{x^4} ).
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 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 1-lnx divided by 2x?
-1
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 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
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 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 4th derivative of lnx?
The fourth derivative of ( \ln(x) ) can be determined by first calculating its derivatives. The first derivative is ( \frac{1}{x} ), the second derivative is ( -\frac{1}{x^2} ), the third derivative is ( \frac{2}{x^3} ), and the fourth derivative is ( -\frac{6}{x^4} ). Thus, the fourth derivative of ( \ln(x) ) is ( -\frac{6}{x^4} ).
