Question

tính đạo hàm của mỗi hàm số sau :

a) y=\(\dfrac{1}{\left(x^2-x+1\right)^5}\) ; b) y=\(x^2+x\sqrt{x}+1\) ; c) y=\(\sqrt{\dfrac{x^2+1}{x}}\)

Answer 1

a : \(y=\dfrac{1}{\left(x^2-x+1\right)^5}=\left(x^2-x+1\right)^{-5}\)

\(\Rightarrow y'=-5\left(2x-1\right)\left(x^2-x+1\right)^{-6}=\dfrac{5-10x}{\left(x^2-x+1\right)^6}\)

b: \(y=x^2+x^{\dfrac{3}{2}}+1\Rightarrow y'=2x+\dfrac{3}{2}x^{\dfrac{1}{2}}=2x+\dfrac{3\sqrt{x}}{2}\)

\(y=\sqrt{\dfrac{x^2+1}{x}}=\left(\dfrac{x^2+1}{x}\right)^{\dfrac{1}{2}}\Rightarrow y'=\dfrac{1}{2}\left(\dfrac{x^2+1}{x}\right)'\left(\dfrac{x^2+1}{x}\right)^{\dfrac{-1}{2}}=\dfrac{x^2-1}{2x^2}\times\dfrac{1}{\sqrt{\dfrac{x^2+1}{x}}}=\dfrac{x^2-1}{2x^2\sqrt{\dfrac{x^2+1}{x}}}\)