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)]
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
The anti-derivative of 1/x is ln|x| + C, where ln refers to logarithm of x to the base e and |x| refers to the absolute value of x, and C is a constant.
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.
ln(3) is a constant. If graphed, it would be a horizontal line. Its derivative is zero.
The derivative of ln(10) is 1/10. This is because the derivative of the natural logarithm function ln(x) is 1/x. Therefore, when differentiating ln(10), the derivative is 1/10.
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)]
the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x
y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
The derivative of ln(x) is 1/x. Therefore, by Chain Rule, we get:[ln(10x)]' = 1/10x * 10 = 1/xUsing this method, you can also infer that the derivative of ln(Ax) where A is any constant equals 1/x.
The derivative of ln x is 1/x. Replacing the expression, that gives you 1 / (1-x). By the chain rule, this must then be multiplied by the derivative of (1-x), which is -1. So, the final result is -1 / (1-x).
-1/ln(1-x) * 1/(1-X) or -1/((1-x)*ln(1-x))
The derivative of logx, assuming base 10, is 1/(xln10).
In this case, you need to apply the chain rule. Note that the derivative of ln N = 1/N. In that case we get: f(x) = ln(1 - x) ∴ f'(x) = 1/(1 - x) × -1 ∴ f'(x) = -1/(1 - x)
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
x (ln x + 1) + Constant