# Using Differentials

• May 14th 2012, 01:01 PM
Bashyboy
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.
• May 14th 2012, 02:19 PM
skeeter
Re: Using Differentials
\$\displaystyle y = x^2 - 5x\$

\$\displaystyle dy = (2x-5)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)
• May 14th 2012, 03:49 PM
Bashyboy
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.
• May 14th 2012, 05:05 PM
HallsofIvy
Re: Using Differentials
Telling you that "x= 4" does tell you that "y" refers to "y(4)".