differentials

• Apr 10th 2008, 12:06 PM
Eric08
differentials
Hi everyone,

Im having trouble with this problem

-use differentials to calculate radical 15.4, to 3 correct decimal places.

-if anyone could help it would be much appreciated
• Apr 10th 2008, 12:16 PM
galactus
We can use the formula $f(x_{0}+{\Delta}x)\approx{f(x_{0})+f'(x_{0}){\Delt a}x}$

Let $f(x)=\sqrt{x}, \;\ f'(x)=\frac{1}{2\sqrt{x}}$

$x_{0}=15, \;\ x_{0}+{\Delta}x=15.4$

We can see then that ${\Delta}x=0.4$

$f(15.4)=\sqrt{15}+\frac{1}{2\sqrt{15}}(0.4)=3.9246 2312416$

Compare to $\sqrt{15.4}=3.92428337407$
• Apr 10th 2008, 12:27 PM
Eric08
ok, so for 3 correct decimal places the answer would be 3.925?