# Differentiating twice

• Jun 14th 2008, 10:14 AM
Silver
Differentiating twice
Alright, the question is:

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

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

I'm stuck on differentiatin twice (Worried)

• Jun 14th 2008, 10:53 AM
Silver
No?? (Worried)
• Jun 14th 2008, 11:35 AM
TwistedOne151
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.
• Jun 14th 2008, 12:40 PM
Silver
Quote:

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 (Blush)
edited it now

thanks for pointing it out
• Jun 14th 2008, 01:25 PM
TwistedOne151
Chain rule
Apply the chain rule. You should have gotten from $y=\frac{1}{x^2}$ that
$\Large{\begin{array}{rcl}\frac{dy}{dt}&=&\frac{dy} {dx}\frac{dx}{dt}\\&=&-\frac{2}{x^3}\frac{dx}{dt}\end{array}}$,
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:
$\frac{d^2y}{dt^2}=\frac{d}{dt}\left(-\frac{2}{x^3}\frac{dx}{dt}\right)$.
Use the product rule:
$\frac{d^2y}{dt^2}=-\frac{d}{dt}\left(\frac{2}{x^3}\right)\cdot\frac{d x}{dt}-\frac{2}{x^3}\frac{d}{dt}\left(\frac{dx}{dt}\right )$
$\frac{d^2y}{dt^2}=-\frac{d}{dt}\left(\frac{2}{x^3}\right)\cdot\frac{d x}{dt}-\frac{2}{x^3}\frac{d^2x}{dt^2}$.
Now apply the chain rule to the remaining derivative:
$\frac{d^2y}{dt^2}=-\frac{d}{dx}\left(\frac{2}{x^3}\right)\cdot\frac{d x}{dt}\cdot\frac{dx}{dt}-\frac{2}{x^3}\frac{d^2x}{dt^2}$
$\frac{d^2y}{dt^2}=-\left(-\frac{6}{x^4}\right)\cdot\left(\frac{dx}{dt}\right )^2-\frac{2}{x^3}\frac{d^2x}{dt^2}$
$\frac{d^2y}{dt^2}=\frac{6}{x^4}\left(\frac{dx}{dt} \right)^2-\frac{2}{x^3}\frac{d^2x}{dt^2}$
And you should be able to proceed from here.

--Kevin C.
• Jun 14th 2008, 01:56 PM
Soroban
Hello, Silver!

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

Quote:

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

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

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

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

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

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

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

• Jun 14th 2008, 01:59 PM
Silver
there is definitely a square, but thanks anyway :)