Need help... Calculus (Differentiate)?

a) y=(2√ x + (6)3√ x - 2√ x^3

b) y=(2x^3-4x^2)(3x^5+x^2)

c) y=(3-2x) / (3+2x)

d) y=(sinx) / (2+secx)

Thnks for all the help

2 respuestas

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  • hace 8 años
    Respuesta preferida

    a) Gonna skip this one because I'm not sure what you mean.

    b) Product Rule: dy/dx f(x)g(x) = f '(x)g(x) + f(x)g'(x)

    y = (2x^3 - 4x^2)(3x^5 + x^2)

    y ' = (6x^2 - 8x)(3x^5 + x^2) + (2x^3 - 4x^2)(15x^4 + 2x)

    You can work it out and simplify it if you like.

    c) Quotient Rule: dy/dx f(x)/g(x) = [f '(x)g(x) - f(x)g'(x)] / [g(x)]^2

    y = (3 - 2x) / (3 + 2x)

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

    y ' = [-6 - 4x - (6 - 4x)] / (3 + 2x)^2

    y ' = -12 / (3 + 2x)^2

    d) Use Quotient Rule again.

    y = sin(x) / (2 + sec(x))

    y ' = [cos(x)(2 + sec(x)) - sin(x)(sec(x)tan(x))] / (2 + sec(x))^2

    y ' = [2cos(x) + 1 - tan^2(x)] / (2 + sec(x))^2 <----- sec = 1/cos

  • hace 4 años

    use the chain rule. take the derivative of the exterior function on an identical time as leaving the interior on my own and then take the derivative of the interior function. derivative of ln(x) = a million/x, so derivative of ln(3x-4) is a million/(3x-4) * 3 = 3 / (3x-4)

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