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I really need help for my calc 1 work pls help i need it by tomorrow afternoon and I'm completely lost. i really hope someone helps.

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(Based on todays review)

  • wizard123

    Quick note on derivatives:
    \frac{d}{dx} x^n = nx^{n-1}

    1)  Use quotient Rule
    (\frac{f}{g})' = \frac{f'g - fg'}{g^2}
    in this case f' = g' = 2x
    = \frac{2x(x^2+c^2) - 2x(x^2-c^2)}{(x^2+c^2)^2} \\  \\ = \frac{4c^2 x}{(x^2 +c^2)^2}


    2)  
        I) h = f-g
    h(x) = x^2 -x- x^{-1}
    h'(x) = 2x -1 - (-x^{-2}) \\  \\ h'(1) = 2 - 1+1 = 2
       II) h = f*g
    h(x) = \frac{x^2-x}{x} = x-1 \\  \\ h'(x) = 1, h'(1) = 1
       III)  h = f/g
    h(x) = \frac{x^2-x}{\frac{1}{x}} = x(x^2-x) = x^3 - x^2 \\  \\ h'(x) = 3x^2 -2x \\  \\ h'(1) = 3-2 = 1

    The answers are 2,1,1. So the order from least to greatest is II,III,I

    3) The equation is showing the limit definition of the derivative except the numerator is using f'(x). This means it is the change in derivative over change in 'x'.  This is definition of 2nd derivative.
    You need to find 2nd derivative of f(x).
    First find 1st derivative:
    f'(x) = 3x^2 +4x-5
    Now take derivative of this new function:
    f''(x) = 6x+4 \\  \\ f''(-3) = -18+4 = -14
    The answer is -14.

    4)
    Before you can differentiate, you need a function y(x)
    This can be done with substitution.
    First solve for 'y' in 2nd equation:
    y = \frac{1}{f(x) + 1}
    Then sub in 1st equation for f(x)
    y = \frac{1}{(\frac{x}{x+1}) + 1} \\  \\ y = \frac{1}{\frac{x+x+1}{x+1}} \\  \\ y = \frac{x+1}{2x+1}
    Now you can take derivative using quotient rule:
    \frac{dy}{dx} = \frac{(1)(2x+1) - 2(x+1)}{(2x+1)^2} \\  \\ \frac{dy}{dx} = \frac{-1}{(2x+1)^2}