y=2(2x+x)2y = 2(2x + \sqrt{x})^{2}y=2(2x+x)2
Let u=2x+xu = 2x + \sqrt{x}u=2x+x
y=2u2y = 2u^{2}y=2u2
dydu=4u\frac{dy}{du} = 4ududy=4u
dudx=2+12x\frac{du}{dx} = 2 + \frac{1}{2\sqrt{x}}dxdu=2+2x1
∴dydx=(dydu)(dudx)\therefore \frac{dy}{dx} = \left( \frac{dy}{du} \right) \left( \frac{du}{dx} \right)∴dxdy=(dudy)(dxdu)
= 4u(2+12x)4u \left( 2 + \frac{1}{2\sqrt{x}} \right)4u(2+2x1)
= 4(2x+x)(2+12x)4 \left( 2x + \sqrt{x} \right) \left( 2 + \frac{1}{2\sqrt{x}} \right)4(2x+x)(2+2x1)
