# Math Help - Differentiation Problem

1. ## Differentiation Problem

So here's the problem:

A curve is represented parametrically by x = (t²-1)², y = t³. Find dy/dx in terms of t and show that
$
\frac {d^2 y}{dx^2} =
\frac {-3(t^2 +1)}{16t(t^2- 1)^3}
$

I tried to work it out, and I got

dy/dx = $
\frac{3t^2}{4t^3-4t}
$
(not sure if it's correct). And about the second derivative, it got too complicated.

Thanks for any help.

2. Originally Posted by fdylan
First of all I'm not sure whether it's in the correct Section.

So here's the problem:

A curve is represented parametrically by x = (t²-1)², y = t³. Find dy/dx in terms of t and show that
$
\frac {d^2 y}{dx^2} =
\frac {-3(t^2 +1)}{16t(t^2- 1)^3}
$

I tried to work it out, and I got

dy/dx = $
\frac{3t^2}{4t^3-4t}
$
(not sure if it's correct). And about the second derivative, it got too complicated.

Thanks for any help.
$x = (t^2-1)^2 \: \rightarrow \: \frac{dx}{dt} = 4t(t^2-1) = 4t^3-4t$

$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{4t(t^2-1)}$ (this will be used later in the chain rule)

$y = t^3 \: \rightarrow \: \frac{dy}{dt} = 3t^2$

The chain rule says that $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$

$\frac{dy}{dx} = \frac{3t^2}{2t(t^2-1)} = \frac{3t}{4t^2-4}$ I have expanded the brackets to make the next step easier

No problem with your original answer although you can cancel out a t.

For the second part use the quotient rule.

$u = 3t \: \rightarrow u' = 3$

$v = 4t^2-4 \: \rightarrow v' = 8t$

3. Many Thanks. Cutting out 't' makes it a lot easier I guess.

4. Originally Posted by fdylan
for the x part, I worked it out different.
$
x = (t^2 -1)^2
x = (t^2 -1)(t^2 -1)
x = t^4 -2t^2 +1
\frac{dx}{dt} = 4t^3 -4t

\frac{dy}{dx} = \frac{3t^2}{4t^3 -4t}
$
It still works, I just used the chain rule.

$\frac{3t^2}{4t^3-4t} = \frac{3t^2}{4t(t^2-1)} = \frac{3t}{4(t^2-1)}$

t cancels