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}{√(x)}\implies y=\cfrac{x^3-2x+1}{x^{(1)/(2)}} \\\\\\ \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{(3x^2-2)√(x)~~~-~~~(x^3-2x+1)\cdot (1)/(2)x^{-(1)/(2)}}{(√(x))^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(3x^2-2)√(x)~~~-~~~(x^3-2x+1)\cdot (1)/(2)\cdot (1)/(√(x))}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(3x^2-2)√(x)~~~-~~~(x^3-2x+1)\cdot(1)/(2√(x))}{x}`


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


`\bf \cfrac{dy}{dx}=\cfrac{(6x^3-4x~~-~~x^3+2x-1)/(2√(x))}{x}\implies \cfrac{dy}{dx}=\cfrac{(5x^3-2x-1)/(2√(x))}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x√(x)}`


`\bf \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^1\cdot x^{(1)/(2)}}\implies \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^{1+(1)/(2)}}\\\\\\ \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^{(3)/(2)}} \implies \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2√(x^3)}`