Find the equation of the tangent to the curve \(y = \frac{x - 1}{2x + 1}, x \neq -\frac{1}{2}\) at the point (1, 0).
\(y = \frac{x - 1}{2x + 1}, x \neq -\frac{1}{2}\)
Using the quotient rule, we have
\(\frac{\mathrm d y}{\mathrm d x} = \frac{(2x + 1). 1 - (x - 1). 2}{(2x - 1)^{2}}\)
= \(\frac{2x + 1 - 2x + 2}{(2x + 1)^{2}}\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{3}{(2x + 1)^{2}}\)
At (1, 0), \(\frac{\mathrm d y}{\mathrm d x} = \frac{3}{(2(1) + 1)^{2}} = \frac{3}{9} = \frac{1}{3}\)
Equation : \(\frac{y - 0}{x - 1} = \frac{1}{3}\)
\(y = \frac{1}{3}(x - 1)\)
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