Maths Question:
$\text{Differentiate }y=\sqrt{x}\text{ from the first principle}$
Maths Solution:
$\begin{align} & y=\sqrt{x} \\ & y+\delta y=\sqrt{x+\delta x} \\ & \delta y=\sqrt{x+\delta x}-\sqrt{x} \\ & \delta y=\sqrt{x+\delta x}-\sqrt{x}\times \frac{\sqrt{x+\delta x}+\sqrt{x}}{\sqrt{x+\delta x}+\sqrt{x}} \\ & \delta y=\frac{x+\delta x-x}{\sqrt{x+\delta x}+\sqrt{x}} \\ & \text{Divide through by }\delta x \\ & \frac{\delta y}{\delta x}=\frac{\delta x}{\delta x\left[ \sqrt{x+\delta x}+\sqrt{x} \right]}=\frac{1}{\sqrt{x+\delta x}+\sqrt{x}} \\ & \text{Proceeding to their limits as }\delta x\to 0 \\ & \frac{dy}{dx}=\underset{\delta x\to \infty }{\mathop{\lim }}\,\frac{\delta y}{\delta x}=\frac{1}{\sqrt{x+0}+\sqrt{x}}=\frac{1}{2\sqrt{x}} \\\end{align}$
University mathstopic:
