# Using differentials to approximate a function

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.

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Looking at it again I realized this $$df=\frac{x}{\sqrt{x^2-y^2}}dx-\frac {y}{\sqrt{x^2-y^2}}dy$$ With $x=5,\Delta x=-.02,y=3,\Delta y=.03$ $$f(x+\Delta x,y+\Delta y)=\sqrt{(5-.02)^2-(3+.03)^2}$$ $$df=\frac{5}{\sqrt{5^2-3^2}}(-.02)-\frac {3}{\sqrt{5^2-3^2}}(.03)$$ $$df=-.0475$$ $$df\approx\Delta f$$ $$f(x+\Delta x,y+\Delta y)=\Delta f+f(x,y)$$ $$f(x+\Delta x,y+\Delta y)\approx3.9525$$
So, in your example $x=5,\ \Delta x=-.02$, and I guess $y=3$ and $\Delta y=.03$. So, try with $y=3$.
Aand.. you should clarify the question. Maybe they want only to compute $\sqrt{(4.98)^2-(3.03)^2} - \sqrt{5^2-3^2}$.