Question

Differentiate with respect to x

Answer 1

`\bf y = \cfrac{6x^2-9x^4}{3x}\implies \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(12x-36x^3)(3x)-(6x^2-9x^4)3}{(3x)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(36x^2-108x^4)~~~~-~~~~(18x^2-27x^4)}{(3x)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{36x^2-108x^4~~~~-18x^2+27x^4}{(3x)^2}\implies \cfrac{dy}{dx}=\cfrac{18x^2-81x^4}{9x^2}`

`\bf \cfrac{dy}{dx}=\cfrac{~~\begin{matrix} 9x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(2-9x^2)}{~~\begin{matrix} 9x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies \cfrac{dy}{dx}=2-9x^2`

Answer 2

The answer is 2x - 3x^3. You have to divide the coefficients and subtract the exponents.