Question

Tính đạo hàm các hàm số sau:
a) \(y=\sqrt{x}+\dfrac{1}{\sqrt{x}}+0,1x^{10}\)
b) \(y=\dfrac{2x^2+x+1}{x^2-x+1}\)
c) \(y=\left(1+3x+5x^2\right)^4\)

Answer 1

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+\dfrac{x^{10}}{10}=U+V+T\)

\(\left\{{}\begin{matrix}U^2=x;\\V^2=\dfrac{1}{x}\\Y'=U'+V'+T'\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\)

\(\left(1\right)\Leftrightarrow U'=\dfrac{1}{2U}=\dfrac{1}{2\sqrt{x}}\)

(2) \(\Leftrightarrow V'=\dfrac{-1}{x^2.2V}=\dfrac{-1}{2x^2.\dfrac{1}{\sqrt{x}}}=\dfrac{-1}{2.\sqrt[3]{x^2}}\)

\(\left(3\right)\Leftrightarrow Y'=\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2\sqrt[3]{x^2}}+x^9\)