If x=t^3, y=2t^2 for t>0, then what does d^2y/dx^2 equal?
This is what the question is asking.
$\displaystyle \frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$
Step 1. Express $\displaystyle y' = \frac{{dy}}{{dx}}$ in terms of t.
Step 2. find $\displaystyle \frac{{dy'}}{{dt}}$, i.e. the derivative of step 1 with respect to t.
Step 3. divide $\displaystyle \frac{{dy'}}{{dt}}$ by $\displaystyle \frac{{dx}}{{dt}}$.
So is this what I should do?
$\displaystyle y=\frac{{2x}}{{t}}
$
Step 1:$\displaystyle
y'=\frac{{2}}{{t}}
$
Step 2: $\displaystyle
\frac{{dy'}}{{dt}}=\frac{{-2}}{{t^2}}
$
Step 3: $\displaystyle
\frac{{dx}}{{dt}}=3t^2
$
so the answer would be $\displaystyle \frac{{d^2y}}{{dx^2}}=\frac{{-2}}{{3t^4}}
$.
Is that right?
first find $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}$
you should get $\displaystyle \frac{dx}{dt} = 3t^2$ and $\displaystyle \frac{dy}{dt} = 4t$
then you use $\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$
giving $\displaystyle \frac{dy}{dx} = 4t \cdot \frac{1}{3t^2}$
want to finish this off ?
Okay, so $\displaystyle
\frac{dy}{dx} = \frac{4}{3t}
$
Step 2: $\displaystyle
\frac{{dy'}}{{dt}}=\frac{{-4}}{{3t^2}}
$
Step 3: $\displaystyle
\frac{{dx}}{{dt}}=3t^2
$
so the answer would be $\displaystyle
\frac{{d^2y}}{{dx^2}}=\frac{{-4}}{{9t^4}}
$.
Is that right?