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
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 $\displaystyle \sqrt{15.4}=3.92428337407$