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
To find dy/dx for the equation y = log(x), we can use the chain rule of differentiation.
First, let's rewrite the equation in exponential form: x = e^y.
Now, differentiate both sides of the equation with respect to x:
d(x)/d(x) = d(e^y)/d(x).
The left side simplifies to 1, and on the right side, we need to apply the chain rule.
Recall that the chain rule states that if u = f(g(x)), then du/dx = f'(g(x)) * g'(x).
In our equation, u = e^y and g(x) = y. So we can apply the chain rule as follows:
1 = e^y * dy/dx.
Now, we solve for dy/dx:
dy/dx = 1 / e^y.
Since x = e^y, we can substitute in x for e^y:
dy/dx = 1 / x.
Therefore, the derivative of y = log(x) is dy/dx = 1 / x.