Question

how to find the derivative of this function?

I keep getting the answer wrong, could someone pls explain this step by step?

thank you

Answer 1

$\bf y=\cfrac{x^3-2x+1}{\sqrt{x}}\implies y=\cfrac{x^3-2x+1}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot \frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x})^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot\frac{1}{2\sqrt{x}}}{x}$

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

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

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