find dy/dx where y = a^x
First, we need to re-write it as e to the power something. The definition of log base e is that e^log(y) = y. We can put our expression into this equality too. So a^x = e^log(a^x), so we use log rules to bring the x down from a power to being at the front of the log, so a^x = e^(x*log(a)).
Now that we are differentiating something in the form e to the power something, we can use standard differentiation to carry it out. When y = e^bx, dy/dx = be^bx, and this is all we need now. So for us, y = e^( log(a) * x ). This means that dy/dx is log(a) * e^( log(a) * x ).
AS
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