Calculus (3rd Edition)

We use the quotient rule to find the derivative of $\frac{x^{4}-4}{x^{2}-5}$ at $x=2$. $\frac{d}{dx}(\frac{x^{4}-4}{x^{2}-5})=\frac{(x^{2}-5)\times\frac{d}{dx}(x^{4}-4)-(x^{4}-4)\times\frac{d}{dx}(x^{2}-5)}{(x^{2}-5)^{2}}$ $=\frac{(x^{2}-5)4x^{3}-(x^{4}-4)2x}{(x^{2}-5)^{2}}=\frac{2x^{5}-20x^{3}+8x}{(x^{2}-5)^{2}}$ $\frac{dy}{dx}|_{x=2}=\frac{2(2)^{5}-20(2)^{3}+8(2)}{((2)^{2}-5)^{2}}=-80$