Toán đạo hàm của hàm số Y=(x^3 – 5). $\sqrt[n]{X}$ bằng biểu thức nào 28/11/2021 By Reese đạo hàm của hàm số Y=(x^3 – 5). $\sqrt[n]{X}$ bằng biểu thức nào
$y=(x^3-5).\sqrt{x}$ $y’=(x^3-5)’\sqrt{x}+(x^3-5).(\sqrt{x})’$ $=3x^2.\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}$ $=\dfrac{6x^3+x^3-5}{2\sqrt{x}}$ $=\dfrac{7x^3-5}{2\sqrt{x}}$ Trả lời
Ta có $y = (x^3 – 5) . \sqrt[n]{x} = \sqrt[n]{x . x^{3n}} – 5\sqrt[n]{x} = x^{\frac{3n+1}{n}} – 5 . x^{\frac{1}{n}}$ Vậy $y’ = \left( x^{\frac{3n+1}{n}} – 5 . x^{\frac{1}{n}} \right)’$ $= \dfrac{3n+1}{n} x^{\frac{2n+1}{n}} – \dfrac{5}{n} x^{\frac{1-n}{n}}$ $= \dfrac{3n+1}{n} \sqrt[n]{x^{2n} .x} – \dfrac{5}{n} \sqrt[n]{x. x^{-n}}$ $= \sqrt[n]{x} \left( \dfrac{3n+1}{n} x^2 – \dfrac{5}{nx} \right)$ Do đó $y’ = \sqrt[n]{x} \left( \dfrac{3n+1}{n} x^2 – \dfrac{5}{nx} \right)$ Trả lời
$y=(x^3-5).\sqrt{x}$
$y’=(x^3-5)’\sqrt{x}+(x^3-5).(\sqrt{x})’$
$=3x^2.\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}$
$=\dfrac{6x^3+x^3-5}{2\sqrt{x}}$
$=\dfrac{7x^3-5}{2\sqrt{x}}$
Ta có
$y = (x^3 – 5) . \sqrt[n]{x} = \sqrt[n]{x . x^{3n}} – 5\sqrt[n]{x} = x^{\frac{3n+1}{n}} – 5 . x^{\frac{1}{n}}$
Vậy
$y’ = \left( x^{\frac{3n+1}{n}} – 5 . x^{\frac{1}{n}} \right)’$
$= \dfrac{3n+1}{n} x^{\frac{2n+1}{n}} – \dfrac{5}{n} x^{\frac{1-n}{n}}$
$= \dfrac{3n+1}{n} \sqrt[n]{x^{2n} .x} – \dfrac{5}{n} \sqrt[n]{x. x^{-n}}$
$= \sqrt[n]{x} \left( \dfrac{3n+1}{n} x^2 – \dfrac{5}{nx} \right)$
Do đó
$y’ = \sqrt[n]{x} \left( \dfrac{3n+1}{n} x^2 – \dfrac{5}{nx} \right)$