Best Answer

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.

Q: What is the derivative of lnx?

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-1/x2

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

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]

ln(√x)=√(lnx) because √x = x^(1/2), ln(x^(1/2))=√(lnx) using a logarithmic property, we can say that .5(lnx)=√(lnx) now, pretend that lnx=y .5y=√y square both sides .25y^2=y subtract y from both sides .25y^2 -y=0 factor y(.25y - 1)=0 so either y=0 or .25y -1 =0 If .25y -1=0, then y=4 so lnx=0 or lnx=4 lnx cannot equal zero because lnx=0 means e^x=0 and that is impossible. Now, we are left with lnx=4 Isolate x by making both sides of the equation powers of e: e^(lnx)=e^4 x=e^4, which is approximately 54.6 Lastly, check this answer by plugging e^4 back into the original equation: ln(√(e^4))=√(ln(e^4)) ln(e^2)=√(4(lne)) 2lne=2√1 2(1)=2 2=2 There you go!

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

-1/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

d/dx lnx=1/x

x (ln x + 1) + Constant

(xlnx)' = lnx + 1

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 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

d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)