y'(x) = 2(1/x^3 - x) * (-3/x^4 - 1)
y'(2) = 2(1/8 - 2)(-3/16 - 1)
= 2(-15/8)(-19/16)
= 285/64
Write the equation of the line tangent to the graph of the function at the indicated point. As a check, graph both the function and the tangent line you found to see whether it looks correct.
(1/x^3 -x)^2 at x= 2
1 answer