Asked by Evan
Given y=log(x), what is dy/dx?
If you have to derive this from first principles, then you have to use the defintion of the Logarithm. E.g. you could use that the Logarithm is the inverse of the exponential function and that the derivative of the exponential function is the exponential function itself.
y = Log(x) -->
x = Exp(y)
dx/dy = Exp(y) = x --->
dy/dx = 1/(dx/dy) = 1/x
If you have to derive this from first principles, then you have to use the defintion of the Logarithm. E.g. you could use that the Logarithm is the inverse of the exponential function and that the derivative of the exponential function is the exponential function itself.
y = Log(x) -->
x = Exp(y)
dx/dy = Exp(y) = x --->
dy/dx = 1/(dx/dy) = 1/x
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