estimate (.99e^.02)^8 with linear approximation.
I'm getting lost trying to apply the formula fx (a,b) delta x + fy (a,b) delta y
I know that e^0 = 1 but I get confused when trying to apply this to the f(x) function to take the derivative of it. so it would be F(x) = (.99e^x)^8 take the deriv of this?
Consider the function \((xe^y)^8\) and you want to estimate that near (1,0)
ohhh, so would I take partial derivs? or am I think of the wrong thing. Because I believe using that formula i menitoned before it delta y =.01 delta x=.02
Yes, you take the partial derivative and delta y = -.01
To make it nicer you probably want to write the function as \(x^8e^{8y}\)
i figured it out! did the easy derivative wrong haha
:) it happens
\[(1+((8(1)^7)*(-.01)) + (0+8e^0)*(.02))\] which =1.08 which is a much better percentage of error.
quaint pre-calculator problem from math teachers youth
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