Q: What is the derivative of natural log?

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Derivative of natural log x = 1/x

The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).

d/dx (2 log(1) + x) = 1

Ever heard of calculator?? log to base 10 = 0.0367087, natural log, 0.08452495

ln(ln)

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Derivative of natural log x = 1/x

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.

2 log(x)derivative form:d/dx(2 log(x)) = 2/x

The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).

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)).

Assuming that is the natural logarithm (logarithm to base e), the derivative of ln x is 1/x. For other bases, the derivative of logax = 1 / (x ln a), where ln a is the natural logarithm of a. Natural logarithms are based on the number e, which is approximately 2.718.

The derivative of ln x, the natural logarithm, is 1/x.Otherwise, given the identity logbx = log(x)/log(b), we know that the derivative of logbx = 1/(x*log b).ProofThe derivative of ln x follows quickly once we know that the derivative of ex is itself. Let y = ln x (we're interested in knowing dy/dx)Then ey = xDifferentiate both sides to get ey dy/dx = 1Substitute ey = x to get x dy/dx = 1, or dy/dx = 1/x.Differentiation of log (base 10) xlog (base 10) x= log (base e) x * log (base 10) ed/dx [ log (base 10) x ]= d/dx [ log (base e) x * log (base 10) e ]= [log(base 10) e] / x= 1 / x ln(10)

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

The natural log of a number is some other number such that if you take e (2.718281828...) and raise it to that other number you would get the first number. Another way to say this is that a natural log is a log with base e. The common log of a number is some other number such that if you take 10 and raise it to that other number you would get the first number. The natural log base, e, is a special transcendental number, chosen so that the derivative (respect to x) of ex is equal to ex . In other words, the slope of a tangent line to the curve y = ex at point (x, ex) is equal to ex for all x

To make a natural log a log with the base of 10, you take ten to the power of you natural log. Ex: ln15=log10ln15=log510.5640138 I'm sorry if you don't have a calculator that can do this, but this will work.

The inverse of the natural log function lnx is exA function must be one to one to have an inverse and the log function is.I am not sure if that is what you are asking.The derivative of ex is itself.That is to say if f(x)=ex then f'(x)=exIf you are asking about the derivative of lnx, it is 1/xand if you look at logb x=1/(xlnb)Not sure which one you are looking for.

Natural log Common log Binary log