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d/dx of lnx is 1/x
Therefore the derivative is 1/(1+x)
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Q: What is the derivative of ln 1 plus x?
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What is the derivative of lnlnx?

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)]


What is the derivative of Ln 2 plus x?

The order of operations is not quite clear here.If you mean (ln 2) + x, the derivate is 0 + 1 = 1.If you mean ln(2+x), by the chain rule, you get (1/x) times (0+1) = 1/x.


How do you derivative x plus y - 1 equals ln x2 plus y2?

There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).


What is the anti derivative of 1 over 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.


What is the derivative of sinx pwr cosx?

For the function: y = sin(x)cos(x) To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm: ln(y) = ln(sin(x)cos(x)) by the properties of logarithms, this can also be expressed as: ln(y) = cos(x)ln(sin(x)) deriving both sides of the equation yields: (1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x)) This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y'). solving for y' in the above, the following is found: y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))] y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.

Related questions

What is the Derivative of 500 ln x plus 1?

the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x


What is the derivative of lnlnx?

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)]


What is the derivative of e the the power ln 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


What is the derivative of 2lnx?

The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x


What is the derivative of ln 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).


What is the derivative of lnln1-x?

-1/ln(1-x) * 1/(1-X) or -1/((1-x)*ln(1-x))


What is the derivative of Ln 2 plus x?

The order of operations is not quite clear here.If you mean (ln 2) + x, the derivate is 0 + 1 = 1.If you mean ln(2+x), by the chain rule, you get (1/x) times (0+1) = 1/x.


What is the derivative of ln 1-x?

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)


How do you derivative x plus y - 1 equals ln x2 plus y2?

There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).


What is the derivitive of ln10x?

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.


What is the anti derivative of lnx?

x (ln x + 1) + Constant


What is the anti derivative of 1 over 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.