How to evaluate x=0 and dx=-.03? I got(1/9)e^(x/9)dx as dy of e^(x/9)and I don't know how I should plug them in.

Answer 1

eh? Just plug them in.

dy = (1/9) e^(x/9) dx

at x=0, dx=-.03, we have

dy = (1/9)(1)(-.03) = -1/300

Answer 2

What his means is that if you draw the graph of y=e^(x/9) and then draw the tangent line at x=0, if you move along that line horizontally to the left by .03 (dx), you will move vertically down by 1/300 (dy).