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Math Help - Help! functions and derivatives

  1. #1
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    Help! functions and derivatives

    Hi,

    Q1.
    For f(x) = 2x^3 + 5x^2 - 7x -4 find f '(x) using the limit definition of the derivative.

    Q2.
    Find the derivatives of f(x) = cos(4x) + sin(5x) up to the fourth order, f '(x), f ''(x), f '''(x) and f ''''(x).

    Confused on these questions - i know derivatives, but not that advanced,

    thanks

    Moon
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  2. #2
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    I haven't done limits in a long time, so I'm not sure on how to go about solving the first one.

    However, for f(x) = cos(4x) + sin(5x) the derivatives are not too hard. As sin and trig are periodic functions, the derivatives will alternate in a pattern.
    For example, if f(x) = cosx, f'(x)=-sinx, f''(x)=-cosx, f'''=sinx, f''''=cosx.

    So he derivatives would be:
    f'(x) = -4sin(4x) + 5cos(5x)
    f''(x) = -16cos(4x) - 5sin(5x)
    f'''(x) = 64sin(4x) - 25cos(5x)

    Then can you figure out the fourth one on your own?
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  3. #3
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    Quote Originally Posted by Moon Hoplite View Post
    Hi,

    Q1.
    For f(x) = 2x^3 + 5x^2 - 7x -4 find f '(x) using the limit definition of the derivative.

    Q2.
    Find the derivatives of f(x) = cos(4x) + sin(5x) up to the fourth order, f '(x), f ''(x), f '''(x) and f ''''(x).

    Confused on these questions - i know derivatives, but not that advanced,

    thanks

    Moon
    Q1)
    This one is all about doing the algebra.


    \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}
    =\lim_{h\rightarrow 0}\frac{[2(x+h)^3 + 5(x+h)^2 - 7(x+h) -4]-[2x^3 + 5x^2 - 7x -4]}{h}
    ...

    Q2)
    Remember:
    \frac{d}{dx}(\sin x) = \cos x
    \frac{d}{dx}(\cos x) = -\sin x
    \frac{d}{dx}[f(g(x))] =\frac{d}{dx}f(g(x))\cdot \frac{d}{dx}g(x) (Chain Rule)

    I will start you off:
    f(x) = \cos(4x) + \sin(5x)
    f'(x) = -4sin(4x) + 5\cos(5x)
    f''(x) = \frac{d}{dx}(f'(x))= -16\cos(4x) - 25\sin(5x)
    ...

    ----
    Edit: xxlvh beats me to it.
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