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2 years ago

How to find the derivative of this function?

Mathematics

1 answer:

Lelechka [254]2 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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