Question

How to find the derivative of (3x/sqrt(4x^2+1))???

Answer 1

`\bf y=\cfrac{3x}{√(4x^2+1)}\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=\cfrac{3√(4x^2+1)-3x\left[ (1)/(2)(4x^2+1)^{-(1)/(2)}\cdot 8x \right]}{(√(4x^2+1))^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{3√(4x^2+1)-\left[ (24x^2)/(2√(4x^2+1)) \right]}{(√(4x^2+1))^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{((3√(4x^2+1))(2√(4x^2+1))-24x^2)/(2√(4x^2+1))}{(√(4x^2+1))^2} \\\\\\`


`\bf \cfrac{dy}{dx}=\cfrac{6(4x^2+1)-24x^2}{2√(4x^2+1)}\cdot \cfrac{1}{(√(4x^2+1))^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{24x^2+6-24x^2}{2√(4x^2+1)}\cdot \cfrac{1}{(√(4x^2+1))^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{6}{2√((4x^2+1)^3)}`