Part 4/5:
\frac{1}{y} \cdot \frac{dy}{dx}
]
The right-hand side requires the product rule. The derivative of (\sqrt{x}) is (\frac{1}{2\sqrt{x}}) and the derivative of (\ln(x)) is (\frac{1}{x}). Thus:
[
\frac{1}{2\sqrt{x}} \cdot \ln(x) + \sqrt{x} \cdot \frac{1}{x} = \frac{1}{2\sqrt{x}} \cdot \ln(x) + \frac{1}{\sqrt{x}}
]
- Set both derivatives equal and solve for (y'):
[
\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \cdot \ln(x) + \frac{1}{\sqrt{x}}
]
- Rearranging gives us the expression for (\frac{dy}{dx}). Finally, substituting back for (y) allows simplification and elucidation of the derivative.