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Math Help - Chain rule, I think

  1. #1
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    Chain rule, I think

    A particle moves along the x-axis with the velocity dx/dt= f(x). Show that the particle's acceleration is f(x)f '(x).

    The notation looks funny. It usually is dy/dx. This time it's the derivative of x with respect to t.

    Tried chain rule (where f is outside and x is inside), and the product rule.
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  2. #2
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    Quote Originally Posted by Truthbetold View Post
    A particle moves along the x-axis with the velocity dx/dt= f(x). Show that the particle's acceleration is f(x)f '(x).

    The notation looks funny. It usually is dy/dx. This time it's the derivative of x with respect to t.

    Tried chain rule (where f is outside and x is inside), and the product rule.
    Hello,

    you have: \frac{dx}{dt}=f(x).

    As you correctly pointed out the acceleration is the first derivative wrt t of the velocity:

    \frac{d^2x}{dt^2}=f'(x) \cdot \underbrace{\frac{dx}{dt}}_{\text{equals f(x)}}=f(x) \cdot f'(x).
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  3. #3
    Senior Member DivideBy0's Avatar
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    Just curious here, but if you're differentiating x with respect to t, shouldn't it be \frac{\,dt}{\,dx}

    As it is in your first post, \frac{\,dx}{\,dt}=f(x) makes little sense.
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    Quote Originally Posted by DivideBy0 View Post
    Just curious here, but if you're differentiating x with respect to t, shouldn't it be \frac{\,dt}{\,dx}

    As it is in your first post, \frac{\,dx}{\,dt}=f(x) makes little sense.
    Nope, the book says dx/dt= f(x).
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  5. #5
    Senior Member DivideBy0's Avatar
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    I find that confusing
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    Quote Originally Posted by earboth View Post
    Hello,

    you have: \frac{dx}{dt}=f(x).

    As you correctly pointed out the acceleration is the first derivative wrt t of the velocity:

    \frac{d^2x}{dt^2}=f'(x) \cdot \underbrace{\frac{dx}{dt}}_{\text{equals f(x)}}=f(x) \cdot f'(x).
    I don't follow.
    WOuldn't the derivative of f(x) just be f'(x) and not f(x)f'(x)?
    I know the notation has to do with something as you showed the difference of \frac{d^2x}{dt^2}=f'(x) and \frac{dx}{dt}.
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  7. #7
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    its the chain rule.
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  8. #8
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    Quote Originally Posted by Truthbetold View Post
    I don't follow.
    WOuldn't the derivative of f(x) just be f'(x) and not f(x)f'(x)?
    I know the notation has to do with something as you showed the difference of \frac{d^2x}{dt^2}=f'(x) and \frac{dx}{dt}.
    Hello,

    "WOuldn't the derivative of f(x) just be f'(x)" This statement is only true if you derivate wrt x. Now the problem says that you have to derivate wrt t

    As shilz222 pointed out you have to use the chain rule because x is a function of t, that means x isn't a single variable here but a function.
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