dy = (1/9) e^(x/9) dx
at x=0, dx=-.03, we have
dy = (1/9)(1)(-.03) = -1/300
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).
Let y=e^(x/9).
a) Find the differential dy.
dy =
b) Evaluate dy if x=0 and dx=−0.03.
dy =
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