# What is the equation of the tangent line of #f(x) = (e^(x)-1)/(x^2-1)# at #x=2#?

Equation of tangent is

The slope of the tangent is the same as the slope of curve at that point, which is given by the value of derivative of the function at that point.

Hence, equation of tangent is

graph{(9y+(e^2-4)x-5e^2+11)(x^2y-y-e^x+1)=0 [-4.785, 5.215, -1.18, 3.82]}

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To find the equation of the tangent line of f(x) = (e^(x)-1)/(x^2-1) at x=2, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the quotient rule:

f'(x) = [(e^x)(x^2-1) - (e^x)(2x)] / (x^2-1)^2

Next, we substitute x=2 into f'(x) to find the slope of the tangent line at x=2:

f'(2) = [(e^2)(2^2-1) - (e^2)(2(2))] / (2^2-1)^2

Finally, we have the slope of the tangent line at x=2. To find a point on the line, we substitute x=2 into f(x):

f(2) = (e^2-1)/(2^2-1)

Therefore, the equation of the tangent line of f(x) = (e^(x)-1)/(x^2-1) at x=2 is:

y - f(2) = f'(2)(x - 2)

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