The order of operations is not quite clear here.
If you mean (ln 2) + x, the derivate is 0 + 1 = 1.
If you mean ln(2+x), by the chain rule, you get (1/x) times (0+1) = 1/x.
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e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
1/xlnx Use the chain rule: ln(ln(x)) The derivative of the outside is1/ln(x) times the derivative of the inside. 1/[x*ln(x)]
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
Oh, dude, the third derivative of ln(x) is -2/(x^3). But like, who really needs to know that, right? I mean, unless you're planning on impressing your calculus teacher or something. Just remember, math is like a puzzle, except no one actually wants to put it together.