Find the area of the triangle formed from the coordinate axes and the tangent line to the curve y = 5x^-1 -1/5x at the point (5,0).
Solution:
Since the derivative of y with respect to x is
"y^{\\prime}(x)=\\frac{d}{d x}\\left[5 x^{-1}-\\frac{1}{5} x\\right]=\\frac{d}{d x}\\left[5 x^{-1}\\right]-\\frac{d}{d x}\\left[\\frac{1}{5} x\\right]=-5 x^{-2}-\\frac{1}{5}"
the slope of the tangent line at the point (5,0) is "y^{\\prime}(5)=-\\frac{2}{5}" . Thus, the equation of the tangent line at this point is
"y-0=-\\frac{2}{5}(x-5) \\text { or equivalently } \\quad y=-\\frac{2}{5} x+2"
Since the y-intercept of this line is 2 , the right triangle formed from the coordinate axes and the tangent line has legs of length 5 and 2, so its area is "\\frac{1}{2}(5)(2)=5"
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