Chapter 4 - Applications of the Derivative - 4.1 Linear Approximation and Applications - Exercises - Page 172: 15

$$ -0.0267$$

Given $$y=\frac{10-x^{2}}{2+x^{2}}, \quad a=1, \quad d x=0.01 $$ Since \begin{align*} f^{\prime}(x)&=\frac{\left(2+x^{2}\right)(-2 x)-\left(10-x^{2}\right)(2 x)}{\left(2+x^{2}\right)^{2}}\\ &=-\frac{24 x}{\left(2+x^{2}\right)^{2}}\\ f'(1)&=\frac{-24}{9} \end{align*} Then \begin{align*} \Delta y&\approx f'(a) dx\\ &=\frac{-24}{9}(0.01)\\ &= -0.0267\end{align*}