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).
Question 1 Report
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).
Answer Details
Find the tangent to \(y=\dfrac{x-1}{2x+1}\) at \((1,0).\)
Differentiate using the quotient rule with \(u=x-1\ (u'=1)\) and \(v=2x+1\ (v'=2):\)
\[\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(1)(2x+1)-(x-1)(2)}{(2x+1)^2}=\frac{2x+1-2x+2}{(2x+1)^2}=\frac{3}{(2x+1)^2}.\]
Gradient at the point. At \(x=1:\)
\[\frac{dy}{dx}=\frac{3}{(2(1)+1)^2}=\frac{3}{9}=\frac{1}{3}.\]
Equation of the tangent through \((1,0)\) with gradient \(\tfrac13:\)
\[y-0=\frac{1}{3}(x-1)\ \Rightarrow\ 3y=x-1\ \Rightarrow\ x-3y-1=0.\]
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