91Ó°ÊÓ

To find the derivative of the function \(f(x)=(\frac{x+1}{x-1})^2\), we will use the chain rule, which states that if we have a function in the form \(f(g(x))\), then the derivative is given by \[f'(g(x)) \cdot g'(x).\] In this case, let \(g(x) = \frac{x+1}{x-1}\) and \(f(u)=u^2\), where \(u=g(x)\). Now, we will find the derivatives of \(f(u)\) and \(g(x)\): 1. For \(f(u)=u^2\), the derivative with respect to \(u\) is \(f'(u) = 2u\). 2. To find the derivative of \(g(x) =\frac{x+1}{x-1}\), we will use the quotient rule, which states that \[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2},\] where \(u(x) = x + 1\) and \(v(x) = x - 1\). Now, find their derivatives: 1. The derivative of \(u(x)=x+1\) is \(u'(x)=1\). 2. The derivative of \(v(x)=x-1\) is \(v'(x)=1\). Apply the quotient rule for \(g'(x)\): \[g'(x)=\frac{u'v - uv'}{v^2}=\frac{(1)(x-1)-(x+1)(1)}{(x-1)^2}=\frac{x-1-x-1}{(x-1)^2}=\frac{-2}{(x-1)^2}.\] Now, using the chain rule, we have: \[f'(x) = f'(g(x)) \cdot g'(x) = 2 \cdot \frac{x+1}{x-1} \cdot \left(-\frac{2}{(x-1)^2}\right).\]

We are given the point \((3,4)\), so we need to evaluate \(f'(x)\) at \(x=3\) to find the slope of the tangent line at this point: \[f'(3) = 2 \cdot \frac{3+1}{3-1} \cdot \left(-\frac{2}{(3-1)^2}\right) = 2 \cdot \frac{4}{2} \cdot \left(-\frac{2}{4}\right) = 4 \cdot (-1) = -4.\] So, the slope of the tangent line at the given point is -4.

Now that we have the slope of the tangent line (-4) and the point \((3, 4)\), we can use the point-slope form to find the equation of the tangent line. The point-slope form is given by: \[y - y_1 = m(x - x_1),\] where \(m\) is the slope, and \((x_1, y_1)\) is the given point. Plug in the values for \(m\), \(x_1\), and \(y_1\): \[y - 4 = -4(x - 3).\] Now, simplify and solve for \(y\): \[y - 4 = -4x + 12 \Rightarrow y = -4x + 12 + 4.\]

The equation of the tangent line to the graph of the function at the given point is: \[y = -4x + 16.\]