How to get $\frac{d^2y}{dx^2}$ for this parametric equation? I've got to get $\frac{d^2y}{dx^2}$, given the parametric equations: $x=t^6-9$ and $y=t-t^2$
At first, I tried to take the derivatives with respect to $t$, of $x$ and $y$ since $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ which gave me: $\frac{1-2t}{6t^5}$
Next, I differentiated that $\bigg($this assumes $\frac{d^2y}{dx^2}=\frac{d}{dt}\bigg[\frac{dy/dt}{dx/dt}\bigg]$$\bigg)$
which gave: $$\frac{(-2)(6t^5)-(1-2t)(30t^4)}{(6t^5)^2}$$ but this isn't equal to $\frac{d^2y}{dx^2}$
So, my next attempt was to get the 2nd derivatives of $x$ and $y$ $\bigg($this assumes $\frac{d^2y}{dx^2}=\bigg[\frac{d^2y/dt^2}{d^2x/dt^2}\bigg]$$\bigg)$ but this doesn't equal $\frac{d^2y}{dx^2}$ either:
$$\frac{d^2x}{dt^2}=30t^4$$ and $$\frac{d^2y}{dt^2}=-2$$
Next, divide to get:
$$\frac{d^2y}{dx^2}=\frac{-2}{30t^4}$$
But that's wrong too... what am I doing wrong?
 A: Let's take $z = \frac{dy}{dx}$.
Now, $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{dz}{dx} = \frac{dz}{dt}\cdot\frac{dt}{dx} = \dfrac{\dfrac{dz}{dt}}{\dfrac{dx}{dt}}  = \dfrac{\dfrac{d}{dt}(\frac{1-2t}{6t^5})}{\dfrac{d}{dt}(t^6-9)}$$
A: $$\frac{d^2y}{dx^2}=\frac{d}{dx} \frac{dy}{dx} =\frac{d}{dx} \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$
$$=\frac{d}{dx} \frac{1-2t}{6t^5} $$
$$=\frac{dt}{dx} \frac{d}{dt} \frac{1-2t}{6t^5}  $$
$$=\frac{1}{6t^5} \frac16 \frac{d}{dt} (t^{-5}-2t^{-4})  $$
$$=\frac{1}{36t^5}    (-5t^{-6}+8t^{-5})  $$
$$=-\frac{5}{36t^{11}}  +\frac{2}{9t^{10}} $$
A: A single dot on top of a variable (like
x,y) represents differentiation with respect to parameter $t$. Two dots on top represent double derivative  with respect to the parameter.
First convert second derivative of y wrt x  as a formula  in terms of parameter $t$ as independent variable.
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\bigg[\frac{dy/dt}{dx/dt}\bigg]=\frac{d}{dt}\bigg[\frac{dy/dt}{dx/dt}\bigg]\frac{dt}{dx}= \frac{\dot y/\dot x}{\dot x}$$
$$= \frac{\dot x \ddot y-\ddot x \dot y }{\dot x^3} \tag 1$$
Find individual derivatives wrt parameter $t$
$$ \dot x=6t^5, \ddot x =30 t, \dot y =1-2t, \ddot y= -2 ;$$
and then plug them into equn 1).
A: Calculation of the second derivative $\frac{d^{2}y}{dx^{2}}$, when $x=x(t)$ and $y=y(t)$.
The first derivative is:
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
We now need the formula of the derivative of a quotient:
$\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$.
Setting $u=\frac{dy}{dt}$ and $v=\frac{dx}{dt}$, we get:
$\frac{du}{dx}=\frac{d}{dx}(\frac{dy}{dt})=\frac{d^{2}y}{dt^{2}}\frac{dt}{dx}$,
$\frac{dv}{dx}=\frac{d}{dx}(\frac{dx}{dt})=\frac{d^{2}x}{dt^{2}}\frac{dt}{dx}$,
Substituting the quantities in the above formula, we have:
$\frac{\frac{dx}{dt}(\frac{d^{2}y}{dt^{2}}\frac{dt}{dx})-\frac{dy}{dt}(\frac{d^{2}x}{dt^{2}}\frac{dt}{dx})}{(\frac{dx}{dt})^{2}}=$
$=\frac{dt}{dx} \frac{\frac{dx}{dt}(\frac{d^{2}y}{dt^{2}})-\frac{dy}{dt}(\frac{d^{2}x}{dt^{2}})}{(\frac{dx}{dt})^{2}}=$
$=\frac{\frac{dx}{dt}(\frac{d^{2}y}{dt^{2}})-\frac{dy}{dt}(\frac{d^{2}x}{dt^{2}})}{(\frac{dx}{dt})^{3}}$.
In our case $x=t^{6}-9$ and $y=t-t^{2}$, with their respective derivatives:
$\frac{dx}{dt}=6t^{5}$,
$\frac{d^{2}x}{dt^{2}}=30t^{4}$;
$\frac{dy}{dt}=1-2t$,
$\frac{d^{2}y}{dt^{2}}=-2$;
$\frac{d^{2}y}{dx^{2}}=\frac{6t^{5}(-2)-(1-2t)30t^{4}}{(6t^5)^{3}}=$,
$=\frac{8t-5}{36t^{11}}$.
