d/dx ln(1-x)
= 1/(1-x)
d/dx 1/(1-x)
= 1-x * d/dx 1 - 1 * d/dx(1-x)/(1-x^2
= 1-x * 0 - 1 * 0 - 1/(1-x)^2
= -1/(1-x)^2
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The derivative, with respect to x, is -x/sqrt(1-x2)
f(x) = 3x4 - 2x2 + 7f ' (x) = 12x3 - 4xf ' ' (x) = 36x2 - 4
The equation is: ln(1+tx)=tx-(h/g)x^2 BTW
Eight: pH is defined as the negative of the logarithm to the base 10 of the actual value.
integral x/(x-1) .dx = x - ln(x-1) + c where ln = natural logarithm and c = constant of integration alternatively if you meant: integral x/x - 1 .dx = c