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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f(x) = 3x4 - 2x2 + 7 The function f'(x) = 12x^(3) - 4x !st Derivative f''(x) = 36x - 4 2nd derivative. The answer!!!!!
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.
The derivative of the moment generating function is the expectation. The variance is the second derivative of the moment generation, E(x^2), minus the expectation squared, (E(x))^2. ie var(x)=E(x^2)-(E(x))^2 :)
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
pH is minus of logarithm of concentration of hydronium ion
pH of a solution is the negative logarithm of the hydrogen-ion concentration
m
f(x) = 3x4 - 2x2 + 7 The function f'(x) = 12x^(3) - 4x !st Derivative f''(x) = 36x - 4 2nd derivative. The answer!!!!!
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.
The derivative of the moment generating function is the expectation. The variance is the second derivative of the moment generation, E(x^2), minus the expectation squared, (E(x))^2. ie var(x)=E(x^2)-(E(x))^2 :)
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
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
1 divided by x to the third power equals x to the negative third. The derivative of x to the negative third is minus three x to the negative fourth.
If x --> 0+ (x tends to zero from the right), then its logarithm tends to minus infinity. On the other hand, x --> 0- (x tends to zero from the left) makes no sense, at least for real numbers, because the logarithm of negative numbers is undefined.
If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)