Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\dfrac{{\sqrt[3]{{4{x^2} + 8}} - \sqrt {8{x^2} + 4} }}{x}\,\,\,khi\,x \ne 0\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,x = 0\end{array} \right.\). Giá trị của \(f'\left( 0 \right)\) bằng:
Trả lời bởi giáo viên
Ta có:
$\begin{array}{l}f'(0)=\mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{4{x^2} + 8}} - \sqrt {8{x^2} + 4} }}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{4{x^2} + 8}} - 2}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {8{x^2} + 4} - 2}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{4{x^2}}}{{{x^2}\left( {{{\sqrt[3]{{4{x^2} + 8}}}^2} + 2\sqrt[3]{{4{x^2} + 8}} + 4} \right)}} - \mathop {\lim }\limits_{x \to 0} \dfrac{{8{x^2}}}{{{x^2}\left( {\sqrt {8{x^2} + 4} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{4}{{{{\sqrt[3]{{4{x^2} + 8}}}^2} + 2\sqrt[3]{{4{x^2} + 8}} + 4}} - \mathop {\lim }\limits_{x \to 0} \dfrac{8}{{\sqrt {8{x^2} + 4} + 2}} = \dfrac{1}{3} - 2 = - \dfrac{5}{3}\end{array}$
Hướng dẫn giải:
Đạo hàm của hàm số \(y = f\left( x \right)\) tại điểm \(x = {x_0}\) là \(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \dfrac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}}\) (nếu tồn tại).