What is the second derivative of #x/(x^2-4)#?
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"Find the limit as x approaches infinity of #xsin(1/x)#?"
#f’(x) = - (x^2 + 4 ) / (x^2 - 4)^2#
#f”(x) = (2x*(x^2 +12))/(x^2-4)^3#
#f(x) = x / (x^2-4)#
# u = x, du = 1#
#v = (x^2 - 4), dv = 2x#
Quotient rule #(v.du - u.dv) / v^2#
#f’(x) = (((x^2-4).1) - (x*2x)) / (x^2 - 4)^2#
#f’(x) = (x^2 - 4 - 2x^2) / (x^2 - 4)^2#
#f’(x) = - (x^2 + 4 ) / (x^2 - 4)^2#
Similarly, the second derivative is to be done.
#u = -(x^2+4), du = -2x#
#v = (x^2 - 4)^2, dv = 2(x^2-4)*2x = 4x(x^2-4)#
#f”(x) = (((x^2-4)^2* (-2x)) +((x^2+4)* (4x*(x^2-4)))) / (x^2-4)^4#
Taking out #(x^2-4) and simplifying,
#f”(x) = (cancel(x^2-4) (-2x^3 + 8x + 4x^3 + 16x))/(cancel(x^2-4)(x^2-4)^3#
#f”(x) = (2x*(x^2 + 12))/(x^2-4^3#
