Question

How can i differentiate this equation?

Answer 1

`\bf y=\cfrac{2x^2-10x}{√(x)}\implies y=\cfrac{2x^2-10x}{x^{(1)/(2)}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(√(x))~~-~~(2x^2-10x)\left( (1)/(2)x^{-(1)/(2)} \right)}{\left( x^{(1)/(2)} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(√(x))~~-~~(2x^2-10x)\left( (1)/(2√(x)) \right)}{\left( x^{(1)/(2)} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(√(x))~~-~~\left( (2x^2-10x)/(2√(x)) \right)}{x}`

`\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(√(x))~~-~~\left( (2x^2-10x)/(2√(x)) \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ ((4x-10)(√(x))(2√(x))~~-~~(2x^2-10x))/(2√(x))}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(√(x))(2√(x))~~-~~(2x^2-10x)}{2x√(x)}`

`\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x√(x)}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x√(x)} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x√(x)} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x√(x)} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x√(x)}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{√(x)}`