Tính đạo hàm cấp hai của hàm số \(y=\dfrac{5x^2-3x-20}{x^2-2x-3}\) .
\(\dfrac{2\left(7x^3+15x^2-93x+77\right)}{\left(x^2-2x-3\right)^3}\).\(\dfrac{2\left(7x^3-15x^2+93x-77\right)}{\left(x^2-2x-3\right)^3}\).\(\dfrac{2\left(7x^3+15x^2+93x-77\right)}{\left(x^2-2x-3\right)^3}\).\(\dfrac{2\left(7x^3-15x^2-93x+77\right)}{\left(x^2-2x-3\right)^3}\).Hướng dẫn giải:- Viết lại hàm số đã cho dưới dạng chính tắc:
\(y=\dfrac{5x^2-3x-20}{x^2-2x-3}=\dfrac{5\left(x^2-2x-3\right)+7x-5}{x^2-2x-3}=5+\dfrac{7x-5}{x^2-2x-3}=5+\dfrac{4\left(x+1\right)+3\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}\)
\(y=5+4\left(x-3\right)^{-1}+3\left(x+1\right)^{-1}\)
\(y'=-4\left(x-3\right)^{-2}-3\left(x+1\right)^{-2}\)
\(y"=8\left(x-3\right)^{-3}+6\left(x+1\right)^{-3}=2\left(\dfrac{4}{\left(x-3\right)^3}+\dfrac{3}{\left(x+1\right)^3}\right)\)
\(y"=\dfrac{2\left[4\left(x+1\right)^3+3\left(x-3\right)^3\right]}{\left(x-3\right)^3\left(x+1\right)^3}=\dfrac{2\left(7x^3-15x^2+93x-77\right)}{\left(x^2-2x-3\right)^2}\)
Đáp số: \(y"=\dfrac{2\left(7x^3-15x^2+93x-77\right)}{\left(x^2-2x-3\right)^2}\)