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Find the derivative of the function

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f(x)=(x^2+1/x^2)(2x+4)

asked Jul 6, 2013 in CALCULUS by anonymous Apprentice

1 Answer

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f(x) = (x^2 + 1/x^2)(2x + 4)

      = x^2(2x) + x^2(4) + 1/x^2(2x) + 1/x^2(4)    [Since (a+b)(c+d) = ac+ad+bc+bd]

      = 2x^3 + 4x^2 + 2/x + 4/x^2

Differentiating on both sides we get,

f'(x) = d/dx(2x^3) + d/dx(4x^2) +d/dx(2/x) + d/dx(4/x^2)

       = 2*3x^2 + 4*2x + 2(-1/x^2) + 4(-2/x^3)

       = 6x^2 + 8x - 2/x^2 - 8/x^3

Therefore f'(x) = 6x^2 + 8x - 2/x^2 - 8/x^3

answered Jul 6, 2013 by joly Scholar

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