
Using Differentials
The problem is to estimate $\displaystyle y + \Delta{y}$ given that $\displaystyle y = x^2  5x$, x = 4, and $\displaystyle \Delta{x} = 0.01$
I have tried this and found that the approximate change in y is dy = 0.03 and the actual change in y is 0.0301, but I am not sure what to do with this information.

Re: Using Differentials
$\displaystyle y = x^2  5x$
$\displaystyle dy = (2x5)dx$
at $\displaystyle x = 4$ , $\displaystyle y = 4$ and $\displaystyle \Delta y \approx dy = 0.03$
$\displaystyle y + \Delta y \approx y + dy = 4 + 0.03 = 3.97$
(note that y(4.01) = 3.9699 for reference)

Re: Using Differentials
Oh, I see. I wonder why they didn't write $\displaystyle y(4) + \Delta{x}$ or something to that effect; it would have been a little easier for a simpleton like me.

Re: Using Differentials
Telling you that "x= 4" does tell you that "y" refers to "y(4)".