1. ## Differentiating twice

Alright, the question is:

transform $\displaystyle 2x\frac{d^2x}{dt^2} -6(\frac{dx}{dt})^2=x^2-3x^4$

into $\displaystyle \frac{d^2y}{dt^2}+y=3$ by using the substitution $\displaystyle y=\frac{1}{x^2}$

I'm stuck on differentiatin twice

No??

3. ## Question

In the result you're trying to get, are the derivatives of y supposed to be with respect to t, or with respect to x (as you have it written)? I'm pretty sure it should be the former.

--Kevin C.

4. Originally Posted by TwistedOne151
In the result you're trying to get, are the derivatives of y supposed to be with respect to t, or with respect to x (as you have it written)? I'm pretty sure it should be the former.

--Kevin C.
woops, it was supposed to be with respect to t, not x, sorry
edited it now

thanks for pointing it out

5. ## Chain rule

Apply the chain rule. You should have gotten from $\displaystyle y=\frac{1}{x^2}$ that
$\displaystyle \Large{\begin{array}{rcl}\frac{dy}{dt}&=&\frac{dy} {dx}\frac{dx}{dt}\\&=&-\frac{2}{x^3}\frac{dx}{dt}\end{array}}$,
as you said it was the second derivative you had trouble with.
The key is to differentiale both sides of the above result with respect to t:
$\displaystyle \frac{d^2y}{dt^2}=\frac{d}{dt}\left(-\frac{2}{x^3}\frac{dx}{dt}\right)$.
Use the product rule:
$\displaystyle \frac{d^2y}{dt^2}=-\frac{d}{dt}\left(\frac{2}{x^3}\right)\cdot\frac{d x}{dt}-\frac{2}{x^3}\frac{d}{dt}\left(\frac{dx}{dt}\right )$
$\displaystyle \frac{d^2y}{dt^2}=-\frac{d}{dt}\left(\frac{2}{x^3}\right)\cdot\frac{d x}{dt}-\frac{2}{x^3}\frac{d^2x}{dt^2}$.
Now apply the chain rule to the remaining derivative:
$\displaystyle \frac{d^2y}{dt^2}=-\frac{d}{dx}\left(\frac{2}{x^3}\right)\cdot\frac{d x}{dt}\cdot\frac{dx}{dt}-\frac{2}{x^3}\frac{d^2x}{dt^2}$
$\displaystyle \frac{d^2y}{dt^2}=-\left(-\frac{6}{x^4}\right)\cdot\left(\frac{dx}{dt}\right )^2-\frac{2}{x^3}\frac{d^2x}{dt^2}$
$\displaystyle \frac{d^2y}{dt^2}=\frac{6}{x^4}\left(\frac{dx}{dt} \right)^2-\frac{2}{x^3}\frac{d^2x}{dt^2}$
And you should be able to proceed from here.

--Kevin C.

6. Hello, Silver!

Is there a typo? . . . Is there really a square in the equation?
. . Without the square, I can do it (backwards).

Transform .$\displaystyle 2x\frac{d^2x}{dt^2} -6 \frac{dx}{dt} \;=\;x^2-3x^4\;\text{ into }\;\frac{d^2y}{dt^2}+y\;=\;3$
by using the substitution $\displaystyle y=\frac{1}{x^2}$

We have: .$\displaystyle y \;=\;x^{-2}\quad\Rightarrow\quad \frac{dy}{dt} \:=\:-2x^{-3}\frac{dx}{dt}$

Then: .$\displaystyle \frac{d^2y}{dx^2} \;=\;-2x^{-3}\frac{d^2x}{dt^2} + 6x^{-4}\frac{dx}{dt}\quad\hdots\quad\text{ product rule}$

Substitute into: .$\displaystyle \underbrace{\frac{d^2y}{dt^2} }+ \underbrace{y} \;=\;3$

. . . . $\displaystyle \overbrace{-2x^{-3}\frac{d^2}{dt^2} + 6x^{-4}\frac{dx}{dt}} + \;\overbrace{x^{-2}}\;=\;3$

. . $\displaystyle -\frac{2}{x^3}\frac{dx^2}{dt^2} + \frac{6}{x^4}\frac{dx}{dt} \;=\;-\frac{1}{x^2}+3$

Multiply by $\displaystyle -x^4\!:\;\;2x\frac{dx^2x}{dt^2} - 6\frac{dx}{dt} \;=\;x^2-3x^4$

7. there is definitely a square, but thanks anyway