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Đạo hàm Giúp mk 2 câu này nha
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Euphemia
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Đáp án:
\(\begin{array}{l}
1)\quad y’ = x^3 + \dfrac{5}{x^2} – \dfrac{3}{\sqrt x} – \dfrac{7}{x^4}\\
2)\quad y’ = \dfrac{12x^3 + 10x}{\sqrt{2x^2 + 1}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\quad y= \dfrac{x^4}{4} – \dfrac{5}{x} – 6\sqrt x + \dfrac{7}{3x^3}\\
\to y’ = \dfrac14(x^4)’ – 5\left(x^{-1}\right)’ – 6\left(\sqrt x\right)’ + \dfrac{7}{3}(x^{-3})’\\
\to y’ = \dfrac14\cdot4x^3 – 5\cdot (-1).x^{-2} – 6\cdot\dfrac{1}{2\sqrt x} + \dfrac73\cdot (-3).x^{-4}\\
\to y’ = x^3 + \dfrac{5}{x^2} – \dfrac{3}{\sqrt x} – \dfrac{7}{x^4}\\
2)\quad y = (2x^2 + 3)\sqrt{2x^2 + 1}\\
\to y’ = (2x^2 + 3)’\sqrt{2x^2 + 1} + (2x^2 + 3).\left(\sqrt{2x^2 + 1}\right)’\\
\to y’ = 4x\sqrt{2x^2 + 1} + (2x^2 + 3)\cdot\dfrac{(2x^2 + 1)’}{2\sqrt{2x^2 + 1}}\\
\to y’ = 4x\sqrt{2x^2 + 1} + (2x^2 + 3)\cdot\dfrac{2x}{\sqrt{2x^2 + 1}}\\
\to y’ = \dfrac{4x(2x^2 + 1) + 2x(2x^2 + 3)}{\sqrt{2x^2 + 1}}\\
\to y’ = \dfrac{12x^3 + 10x}{\sqrt{2x^2 + 1}}
\end{array}\)
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