let assume i have a position function in 1 dimension with constant acceleration.

$$ x(t) = x_0 + v_0t + \frac{1}{2}at^2 $$

then it's first derivative is a velocity function: $$ \frac{dx}{dt} = v(t) = v_0 + at $$

then it's second derivative is an acceleration function:

$$ \frac{dv}{dt} = a(t) = a $$

so in conclusion if we have x(t) a position function and we take a first derivative, we will get a velocity function and if we take it's second derivative we will get an acceleration function. this is what everyone knows.

now i see some lecture video says that:

$$ \frac{dv}{dt} = \frac{dv}{dx} * \frac{dx}{dt} $$

if this is true then if i calculate $\frac{dv}{dx}$ and multiply by $\frac{dx}{dt}$ i will also get $a(t)$ but i don't know how to do it because when we apply chain rule we need to determine what is inner function and what is outer function but here there is only one function which is x(t) how to find $\frac{dv}{dx}$?

can someone rewrite the position function into inner part and outer part? or what is the valid way to do the calculation?

I'm very new to calculus and physic please explain step by step and easy simple example

  • $\begingroup$ You have an equation for $x$ in terms of $t$ and an equation for $v$ in terms of $t$. Eliminate $t$. $\endgroup$
    – amd
    Commented Nov 24, 2017 at 23:36

2 Answers 2


$$ \frac{dv}{dx} = \frac{d(v_0+at)}{dx} = a\frac{dt}{dx} \tag{$\ast$}$$ Here, $v_0$ and $a$ is assumed to be purely independent constants. To find $\frac{dt}{dx}$, we can either write time in terms of position or differentiate the position function with respect to position which I prefer to do. So, we have $x = x_0+v_0t+\frac{1}{2}at^2$, differentiating both sides with respect to $x$, we get: $$\frac{dx}{dx}=1=\frac{d(x_0+v_0t+\frac{1}{2}at^2)}{dx} = v_0\frac{dt}{dx}+\frac{1}{2}a(2t\frac{dt}{dx})=(v_0+at)\frac{dt}{dx} \\ \implies \frac{dt}{dx}=\frac{1}{v_0+at}$$ Substitute this into $(\ast)$ and multiply with $\frac{dx}{dt}=v(t)=v_0+at$ : $$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}=(\frac{a}{v_0+at})(v_0+at)=a$$

  • $\begingroup$ your answer is so great ^^ very easy to understand. Thank you very much. in your answer you also mentioned about another approch which is writing time in terms of position. can you please also show that? $\endgroup$ Commented Nov 25, 2017 at 18:40
  • 1
    $\begingroup$ @user3270418 The other approach corresponds to treating the position function as a quadratic and solving for t. There will be two solutions but to avoid weird consequences you can take the positive one and differentiate with respect to x. The result may seem different but it gives the same result, just hidden in a square root. As you can see, it seems tiresome and the other approach is much more easier. $\endgroup$ Commented Nov 25, 2017 at 22:03

Note that these equations only hold for constant acceleration, so it's incorrect to write $a = a(t)$.

You can isolate $t$ from the velocity equation

$$ t = \frac{v - v_0}{a} $$

Plugging into the displacement gives

$$ x = x_0 + \frac{v_0(v-v_0)}{a} + \frac{1}{2}\frac{(v-v_0)^2}{a} = x_0 + \frac{v^2 - {v_0}^2}{2a} $$

Solving for $v$ explicitly will yield an ambiguous square root unless we restrict $v > 0$, so instead we can employ implicit differentiation to get

$$ 1 = \frac{v}{a} \frac{dv}{dx} $$

which satisfies

$$ a = v\ \frac{dv}{dx} $$

  • $\begingroup$ in my question the equation has constant acceleration which is a. so a is a constant. so a(t) = a is a valid function. it's like writing f(x) = 3. $\endgroup$ Commented Nov 25, 2017 at 18:30
  • $\begingroup$ am i understand correctly? $\endgroup$ Commented Nov 25, 2017 at 22:19
  • 1
    $\begingroup$ Yes, that's fine. $\endgroup$
    – Dylan
    Commented Nov 25, 2017 at 22:33

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