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Varvara68 [4.7K]
3 years ago
11

How to find the derivative of this function?

Mathematics
1 answer:
Lelechka [254]3 years ago
8 0
\bf y=\cfrac{x^3-2x+1}{\sqrt{x}}\implies y=\cfrac{x^3-2x+1}{x^{\frac{1}{2}}}
\\\\\\
\cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot \frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x})^2}}
\\\\\\
\cfrac{dy}{dx}=\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{x}}}{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot\frac{1}{2\sqrt{x}}}{x}

\bf \cfrac{dy}{dx}=\cfrac{(3x^2-2)\sqrt{x}~~~-~~~\cdot\frac{x^3-2x+1}{2\sqrt{x}}}{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{[(3x^2-2)\sqrt{x}](2\sqrt{x})~~-~~(x^3-2x+1)}{2\sqrt{x}}}{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{[(3x^2-2)2x]~~-~~(x^3-2x+1)}{2\sqrt{x}}}{x}
\implies 
\cfrac{dy}{dx}=\cfrac{\frac{[6x^3-4x]~~-~~(x^3-2x+1)}{2\sqrt{x}}}{x}

\bf \cfrac{dy}{dx}=\cfrac{\frac{6x^3-4x~~-~~x^3+2x-1}{2\sqrt{x}}}{x}\implies 
\cfrac{dy}{dx}=\cfrac{\frac{5x^3-2x-1}{2\sqrt{x}}}{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x\sqrt{x}}

\bf \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^1\cdot  x^{\frac{1}{2}}}\implies \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^{1+\frac{1}{2}}}\\\\\\ \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^{\frac{3}{2}}}
\implies
\cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2\sqrt{x^3}}
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