# Math Help - Can someone explain how this makes 2nd order diff. makes sense

1. ## Can someone explain how this makes 2nd order diff. makes sense

Just reading through a proof which states that:

dy/dt=(-2x^-3)(dx/dt)

Then it states that:

d2y/dt2= (-2x^-3)(d2x/dt2)+(6x^-4)(dx/dt)^2

I can see that the chain rule is used, although I do not understand why the last dx/dt is squared - could somone explain how the first equation transforms to the next please,

Many thanks,

Bryn

2. Originally Posted by Bryn
Just reading through a proof which states that:

dy/dt=(-2x^-3)(dx/dt)

Then it states that:

d2y/dt2= (-2x^-3)(d2x/dt2)+(6x^-4)(dx/dt)^2

I can see that the chain rule is used, although I do not understand why the last dx/dt is squared - could somone explain how the first equation transforms to the next please,

Many thanks,

Bryn
You are differentiating with respect to $t$..

what is the derivative of $-2x^{-3}$? it is $6x^{-4} \frac{dx}{dt}$

that is where the second $\frac{dx}{dt}$ came from..

3. Originally Posted by Bryn
Just reading through a proof which states that:

dy/dt=(-2x^-3)(dx/dt)

Then it states that:

d2y/dt2= (-2x^-3)(d2x/dt2)+(6x^-4)(dx/dt)^2

I can see that the chain rule is used, although I do not understand why the last dx/dt is squared - could somone explain how the first equation transforms to the next please,

Many thanks,

Bryn
$\frac{dy}{dt}=(-2x^{-3}) \frac{dx}{dt}$

Now taking the derivative with respect to t of both sides gives

$\frac{d}{dt} \left( \frac{dy}{dt} \right)= \frac{d}{dt} \left(-2x^{-3}\frac{dx}{dt} \right)$

Now using the product rule on the right side we get

$\frac{d^2y}{dt^2}=\left(\frac{d}{dt}(-2x^{-3}) \right)\cdot \frac{dx}{dt}+(-2x^{-3})\cdot \frac{d}{dt}\left( \frac{dx}{dt}\right)$

Now using the chain rule we get

$\frac{d^2y}{dt^2}=\left((6x^{-4})\cdot \frac{dx}{dt} \right)\cdot \frac{dx}{dt}+(-2x^{-3})\cdot \left( \frac{d^2x}{dt^2}\right)$

$\frac{d^2y}{dt^2}=\left((6x^{-4})\cdot\left( \frac{dx}{dt} \right)^2\right)-(2x^{-3})\left( \frac{d^2x}{dt^2}\right)$

4. Originally Posted by TheEmptySet

$\frac{d^2y}{dt^2}=\left((6x^{-4})\cdot \frac{d^2x}{dt^2} \right)-(2x^{-3})\left( \frac{d^2x}{dt^2}\right)$
beware of these.. $\frac{d^2x}{dt^2} \neq \left(\frac{dx}{dt}\right)^2$