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.