Posted by **Tyler** on Wednesday, June 26, 2013 at 9:37am.

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.

- Please Help with calculus -
**Steve**, Wednesday, June 26, 2013 at 11:42am
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

- Please Help with calculus -
**Steve**, Wednesday, June 26, 2013 at 11:46am
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).

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