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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)).
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∙ 13y ago
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What is the Second derivative of lnx?
-1/x2
What is the derivative of 4 divided by X?
-4/x2
What is the anti-derivative of x2 1x?
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
What is the derivative of square root 2x-x2?
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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 3cosx2?
I'm assuming your question reads "What is the derivative of 3cos(x2)?" You must use the Chain Rule. The derivative of cos(x2) equals -sin(x2) times the derivative of the inside (x2), which is 2x. So... d/dx[3cos(x2)] = -6xsin(x2)
Derivative of natural log?
Derivative of natural log x = 1/x
What is derivative of x2?
2x
What is the derivative of ln1 divided by x?
Given y=ln(1/x) y'=(1/(1/x))(-x-2)=(1/(1/x))(1/x2)=x/x2=1/x Use the chain rule. The derivative of ln(x) is 1/x. Instead of just "x" inside the natural log function, it's "1/x". Since the inside of the function is not x, the derivative must be multiplied by the derivative of the inside of the function. So it's 1/(1/x) [the derivative of the outside function, natural log] times -x-2=1/x2 [the derivative of the inside of the function, 1/x] This all simplifies to 1/x So the derivative of ln(1/x) is 1/x
What is the derivative of x to the second powerr?
2x is the first derivative of x2.
What is the derivative of x to the second power?
2x is the first derivative of x2.
Derivative of log?
The derivative of a log is as follows: 1 divided by xlnb Where x is the number beside the log Where b is the base of the log and ln is just the natural log.
How do you show log x is convex?
Log x is defined only for x > 0. The first derivative of log x is 1/x, which, for x > 0 is also > 0 The second derivative of log x = -1/x2 is always negative over the valid domain for x. Together, these derivatives show that log x is a strictly monotonic increasing function of x and that its rate of increase is always decreasing. Consequently log x is convex.
What is the anti derivative of x squared plus x?
The anti-derivative of X2 plus X is the same as the anti-derivative of X2 plus the anti-derivative of X. The anti derivative of X2 is X3/3 plus an integration constant C1 The anti derivative of X is X2/2 plus an integration constant C2 So the anti-derivative of X2+X is (X3/3)+(X2/2)+C1+C2 The constants can be combined and the fraction can combined by using a common denominator leaving (2X3/6)+(3X2/6)+C X2/6 can be factored out leaving (X2/6)(2X+3)+C Hope that helps
What is the Second derivative of lnx?
-1/x2
What is derivative of 4cosx3?
-12sin(x3)x2
What is the derivative of -2X2?
(-2 x2)' = -4 x
