So I have a homework problem that I cannot figure out. I am supposed to approximate the value of $\sqrt{(4.98)^2-(3.03)^2}$ using differentials. What I have so far is $f(x,y)=\sqrt{x^2-y^2}$ $\Delta f=f(x+\Delta x,y+\Delta y)-f(x,y)$ $df= \frac {\delta f}{\delta x}dx+\frac {\delta f}{\delta y}dy$ Can I do this? $ df=\frac{x}{\sqrt{x^2-y^2}}dx-\frac {y}{\sqrt{x^2-y^2}}dy$ $\sqrt{(5-.02)^2-(4-.97)^2}$ $ df=\frac{5}{\sqrt{5^2-4^2}}(-.02)-\frac {4}{\sqrt{5^2-4^2}}(-.97)$ $df=\frac{-5}{3}(.02)+\frac 43(.97)$ $df\approx 1.26 $
I have the solutions manual and it says the answer should be $3.95$.
What did I do wrong, and how can I get to their answer?
I appreciate any help that anyone has.