Differantiate using first principus. \[ y=\frac{1}{\sqrt{x}} \]

To differentiate the function \( y = \frac{1}{\sqrt{x}} \) using the first principles, we start with the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] In this case, we have \( f(x) = \frac{1}{\sqrt{x}} \). We need to compute \( f(x+h) \): \[ f(x+h) = \frac{1}{\sqrt{x+h}} \] Now, we can substitute \( f(x) \) and \( f(x+h) \) into the derivative formula: \[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \] To simplify the expression in the limit, we can find a common denominator for the two fractions in the numerator: \[ \frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}} = \frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}} \] Thus, we have: \[ f'(x) = \lim_{h \to 0} \frac{\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}}{h} \] This simplifies to: \[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h \sqrt{x} \sqrt{x+h}} \] Next, we can multiply the numerator and denominator by the conjugate of the numerator, \( \sqrt{x} + \sqrt{x+h} \): \[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}{h \sqrt{x} \sqrt{x+h} (\sqrt{x} + \sqrt{x+h})} \] The numerator simplifies as follows: \[ (\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h}) = x - (x+h) = -h \] So we have: \[ f'(x) = \lim_{h \to 0} \frac{-h}{h \sqrt{x} \sqrt{x+h} (\sqrt{x} + \sqrt{x+h})} \] We can cancel \( h \) in the numerator and denominator (as long as \( h \neq 0 \)): \[ f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x} \sqrt{x+h} (\sqrt{x} + \sqrt{x+h})} \] Now, as \( h \to 0 \), \( \sqrt{x+h} \to \sqrt{x} \). Therefore, we can substitute \( \sqrt{x+h} \) with \( \sqrt{x} \): \[ f'(x) = \frac{-1}{\sqrt{x} \cdot \sqrt{x} \cdot (\sqrt{x} + \sqrt{x})} = \frac{-1}{x \cdot 2\sqrt{x}} = \frac{-1}{2x\sqrt{x}} \] Thus, the derivative of \( y = \frac{1}{\sqrt{x}} \) is: \[ \boxed{-\frac{1}{2x\sqrt{x}}} \]