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Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Calculus help... On derivatives. Relatively simple, just confused?

Find the derivative--

f(x) = x [ 1 - (4 / x + 3) ]

Thanks!

4 Answers

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  • 1 decade ago
    Favorite Answer

    f(x) = x[1 - (4 / x + 3)]

    Rewrite the function to make it a bit easier to find the derivative.

    x(1 - (4/x) - 3)

    x(-2 - (4/x))

    f (x) = -2x - 4

    f ' (x) = -2

  • 1 decade ago

    Use the Chain Rule.

    Let g(x) = x, and h(x) = 1 - (4/x + 3).

    The Chain Rule states f(x) = g(x)h(x) -> f'(x) = g'(x)h(x) + g(x)h'(x).

    So, f'(x) = 1- (4/x + 3) + x[8/(x^2)].

    Simplification yields 1-4/x - 3 + 4/x = -2

  • 1 decade ago

    you could use product rule: der of the 1st times the 2nd+der of the 2nd times the first

    but since the first is only a monomial (x) multiply the x through then use power rule

    the second part uses quotient rule:der top times the bottom -der bottom times the top all over the bottom squared

    f(x)=x- 4x/(x+3)

    f '(x)=1- (4(x+30)-1(4x))

    (x+3)^2

    so f '(x)=1- 4x+120-4x =1- 120

    (x+3)^2 (x+3)^2

    just algebra after that

  • 1 decade ago

    f(x) = x [ 1 - (4 / x + 3) ]

    f ' (x) = x [ 0 - ( - 4 / (x + 3)^2 ] + [ 1 - (4 / x + 3) ]

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

    = [ 4x + x^2 + 2x - 3 ] / (x + 3)^2

    = (x^2 + 6x - 3 ) / (x + 3)^2

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