Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

calculus is hard need help!?

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  • 1 month ago
    Favorite Answer

    f(x) = 2x².(3x - 9)

    The function f looks like (u.v), so its derivative looks like: (u'.v) + (v'.u) → where:

    u = 2x² → u' = 4x

    v = 3x - 9 → v' = 3

    f'(x) = (u'.v) + (v'.u)

    f'(x) = 4x.(3x - 9) + (3 * 2x²)

    f'(x) = 12x² - 36x + 6x²

    f'(x) = 18x² - 36x

    f(x) = (3x - 6)/[x.(x + 6)]

    f(x) = (3x - 6)/(x² + 6x)

    The function f looks like (u/v), so its derivative looks like: [(u'.v) - (v'.u)]/v² → where:

    u = 3x - 6 → u' = 3

    v = x² + 6x → v' = 2x + 6

    f'(x) = [(u'.v) - (v'.u)]/v²

    f'(x) = [3.(x² + 6x) - (2x + 6).(3x - 6)]/(x² + 6x)²

    f'(x) = [3x² + 18x - (6x² - 12x + 18x - 36)]/(x² + 6x)²

    f'(x) = [3x² + 18x - (6x² + 6x - 36)]/(x² + 6x)²

    f'(x) = [3x² + 18x - 6x² - 6x + 36]/(x² + 6x)²

    f'(x) = [- 3x² + 12x + 36]/(x² + 6x)²

    f'(x) = 3.(- x² + 4x + 12)/(x² + 6x)²

    f(x) = [(x³ - 1)⁵ + 2]³ → recall the derivative of u^(n) → n * u' * n^(n - 1)

    f'(x) = 3 * [(x³ - 1)⁵ + 2]' * [(x³ - 1)⁵ + 2]² → the derivative of 2 is zero

    f'(x) = 3 * [(x³ - 1)⁵]' * [(x³ - 1)⁵ + 2]²

    f'(x) = 3.[5 * (x³ - 1)' * (x³ - 1)⁴] * [(x³ - 1)⁵ + 2]² → the derivative of 1 is zero

    f'(x) = 3.[5 * (x³)' * (x³ - 1)⁴] * [(x³ - 1)⁵ + 2]²

    f'(x) = 3.[5 * (3x²) * (x³ - 1)⁴] * [(x³ - 1)⁵ + 2]²

    f'(x) = 3.[15x² * (x³ - 1)⁴] * [(x³ - 1)⁵ + 2]²

    f'(x) = 45x².(x³ - 1)⁴.[(x³ - 1)⁵ + 2]²

  • Anonymous
    1 month ago

    As many do, it is an Educational Fraud.

    less than 5% of Students will need let a lone use Calculus in their professional lives.

    Same with Second Languages.

    Stress on students, unnecessary money making classes.

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