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

Printable View

- Apr 10th 2008, 12:06 PMEric08differentials
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 PMgalactus
We can use the formula $\displaystyle f(x_{0}+{\Delta}x)\approx{f(x_{0})+f'(x_{0}){\Delt a}x}$

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

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

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

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

Compare to $\displaystyle \sqrt{15.4}=3.92428337407$ - Apr 10th 2008, 12:27 PMEric08
ok, so for 3 correct decimal places the answer would be 3.925?