Obliczmy pochodną podanej funkcji.

$f^{\prime }\left(x\right)={\left(\frac{25x^2-9}{x^2+2}\right)}^{\prime }=\frac{{\left(25x^2-9\right)}^{\prime }\cdot \left(x^2+2\right)-\left(25x^2-9\right)\cdot {\left(x^2+2\right)}^{\prime }}{{\left(x^2+2\right)}^2}=\frac{50x\left(x^2+2\right)-2x\left(25x^2-9\right)}{{\left(x^2+2\right)}^2}=\frac{50x^3+100x-50x^3+18x}{{\left(x^2+2\right)}^2}=\frac{118x}{{\left(x^2+2\right)}^2}$

Zatem dla argumentu x=10 pochodna ma wartość

$f^{\prime }\left(10\right)=\frac{118\cdot 10}{{\left(10^2+2\right)}^2}=\frac{1180}{{\left(102\right)}^2}=\frac{1180}{10404}=\frac{295}{2601}$