Derivative of 2^x=ln(2)*2x

How does the Derivative of 2^x=ln(2)*2x?

All I can think if is getting (x=)*2^(x-1) as its deriv...

Please help anon...

I'm not sure your method 2 is correct here.

Using y as the dependent:
y=2^x, so to get at x we take the log of both sides:

ln (y) = ln(2^x)
ln y = xln 2
The definition of a log says:
y = e^(xln 2).

differentiating y would then give:
dy/dx = d(e^(xln 2))/dx
which is not yln 2.

I believe method 1 will give proper results.