The point PP lies on the curve y=ln⁡(x+1x−1)y = \ln\left(\frac{x+1}{x-1}\right) for x>1x > 1. The normal to the curve at PP is parallel to the line 2y=3x+22y = 3x + 2. The equation of the tangent to the curve at PP cuts the yy-axis at QQ. Find the area of the triangle POQPOQ, where OO is the origin.

Applications differentiationParallel perpendicularFunctions and graphsStandard derivativesDerivative as gradientArea rectilinear中文
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