Question

$y=\sqrt{(x-3{)}^2+(x^2-2{)}^2}-\sqrt{x^2+(x^2-1{)}^2}$ find $\frac{dy}{dx}$?

Answer 1

$y=\sqrt{{(x-3)}^2+{(x^2-2)}^2}-\sqrt{x^2+{(x^2-1)}^2}$

$\therefore \frac{dy}{dx}=\frac{d}{dx}=\sqrt{{(x-3)}^2+{(x^2-2)}^2}-\sqrt{x^2+{(x^2-1)}^2}$

$\therefore \frac{dy}{dx}=\frac{2(x-3)+2(x^2-2)\left(2x\right)}{2\sqrt{{(x-3)}^2+{(x^2-2)}^2}}-\frac{2x+2(x^2-1)\left(2x\right)}{2\sqrt{x^2+{(x^2-1)}^2}}$

$\therefore \frac{dy}{dx}=\left(\frac{(x-3)+2x^3-4x}{\sqrt{{(x-3)}^2+{(x^2-2)}^2}}\right)-\left(\frac{x+2x^3-2x}{\sqrt{x^2+{(x^2-1)}^2}}\right)$

$\therefore \frac{dy}{dx}=\left(\frac{2x^3-3x-3}{\sqrt{{(x-3)}^2{(x^2-2)}^2}}\right)-\left(\frac{2x^3-x}{\sqrt{x^2{+(x^2-1)}^2}}\right)$