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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?
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.
How do you find second derivative of a function?
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
What is the derivative of log x-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)).
How do you find derivative of x2lnx?
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]
