Answer to Question #340774 in Calculus for Jayson

Question

Solve using differentiation rule

y = 2x ^ 2 * sqrt(2 - x)

Answer 1

$$y'=(2x^2\sqrt{2-x})'=4x\sqrt{2-x}+\dfrac{2x^2}{2\sqrt{2-x}(-1)}$$

$$=4x\sqrt{2-x}-\dfrac{x^2}{\sqrt{2-x}}$$

$$=\dfrac{4x(2-x)-x^2}{2\sqrt{2-x}}=\dfrac{8x-5x^2}{2\sqrt{2-x}}$$