What is the derivative of #y=sqrtx# by first principles?

To find the derivative of ( y = \sqrt{x} ) using first principles:

Let ( f(x) = \sqrt{x} ).

We'll use the definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

Substitute ( f(x) = \sqrt{x} ):

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} ]

To simplify, we'll rationalize the numerator:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} ]

[ f'(x) = \lim_{h \to 0} \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} ]

[ f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})} ]

[ f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} ]

As ( h ) approaches 0, the expression becomes:

[ f'(x) = \frac{1}{2\sqrt{x}} ]

Therefore, the derivative of ( y = \sqrt{x} ) with respect to ( x ) by first principles is:

[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} ]