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the derivative of ln x = x'/x; the derivative of 1 is 0
so the answer is 500(1/x)+0 = 500/x
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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).
Is there a function where the first derivative goes to infinity for x going to 0 and where the first derivative equals 0 when x is 1?
Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.
If y equals log x what is dy over dx?
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.
What is the derivative of X-1 to the power of X?
d/dx (X - 1)x = (X - 1)x ln(X - 1) * x = X(X- 1)x ln(X - 1) -------------------------
What is the derivative of y equals xlnx?
Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x
What is the derivative of ln 1 plus x?
d/dx of lnx is 1/x Therefore the derivative is 1/(1+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 ln 1?
It is equal to 0
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 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 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 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 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 derivative of lnln1-x?
-1/ln(1-x) * 1/(1-X) or -1/((1-x)*ln(1-x))
Derivative of logx?
The derivative of logx, assuming base 10, is 1/(xln10).
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)
