find dy/dx for y=In(x2+5)?

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  • Anonymous
    2 years ago

    y=In(x^2+5)

    d y / d x = [ 2 x / (x^2+5) ]

  • 2 years ago

    y = ln(x^2 + 5)

    dy/dx = 2x/(x^2 + 5)

  • Vaman
    Lv 7
    2 years ago

    y=In(x2+5)

    This gives x2+5 = exp y

    Take the derivative. 2x = exp y dy/dx

    Therefore dy/dx = 2x/exp y= 2x/(x^2 +5)

  • Como
    Lv 7
    2 years ago

    y = ln ( x² + 5 )

    Let u = x² + 5

    du/dx = 2x

    y = ln u

    dy/du = 1/u

    dy/dx = (1/u) (2x) = 2x / ( x² + 5 )

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  • 2 years ago

    assuming you mean y = ln(x^2 + 5)...

    Let u = x^2 + 5

    so y = ln(u)

    dy/du = 1/u

    du/dx = 2x

    by the chain rule dy/dx = dy/du * du/dy = 2x/(x^2 + 5)

  • Anonymous
    2 years ago

    Use chain rule

    u = x^2 + 5

    du/dx = 2x

    dy/du = ln(u)

    dy/du = 1/u

    dy/dx = dy/du * du/dx

    dy/dx = 1/u * 2x

    dy/dx = 2x / (x^2+5)

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