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.
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
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).
To evaluate the expression dy = (1/9)e^(x/9)dx when x = 0 and dx = -0.03, you can follow these steps:
Step 1: Substitute the values of x and dx into the expression dy = (1/9)e^(x/9)dx.
dy = (1/9)e^(0/9)(-0.03)
Step 2: Simplify the expression.
Since e^0 equals 1, we can simplify the equation as follows:
dy = (1/9)(1)(-0.03)
Therefore, dy = -0.003/9.
Step 3: Calculate the result.
To calculate dy, divide -0.003 by 9:
dy = -0.003/9 = -0.0003333 (rounded to six decimal places).
So, when x = 0 and dx = -0.03, dy would be approximately equal to -0.0003333.
Remember, when plugging in values for x and dx, it's essential to substitute them into the right places in the equation and simplify the expression accordingly.