<s>[INST] $y=\frac{1}{x}$, show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find $\frac{d^2y}{dt^2}$ If $y=\frac{1}{x}$, where $x$ is a function of $t$ show that $\frac{dy}{dt}=\frac{-1}{x^2}\frac{dx}{dt}$ and find an expression for $\frac{d^2y}{dt^2}$.
For $\frac{d^2y}{dt^2}$ I keep getting 
$$\frac{d^2y}{dt^2}=\frac{-1}{x^2}\frac{d^2x}{dt^2}+\frac{2}{x^3}\frac{dx}{dt}$$ 
but the correct answer is apparently 
$$\frac{d^2y}{dt^2}=\frac{-1}{x^2}\frac{d^2x}{dt^2}+\frac{2}{x^3}\left({\frac{dx}{dt}}\right)^2$$
 [/INST] $$
\dfrac{d}{dt}g(x)\dot{x}= \dfrac{dg}{dt}\dot{x} + g(x)\ddot{x}
$$
Now let's have $g(x) = -1/x^2$
We have
$$
\dfrac{dg}{dt} = \dfrac{dg}{dx}\dot{x} = \frac{2}{x^3}\dot{x}
$$
So hopefully it is clear to see that you have compute the derivative of $-1/x^2$ wrong?
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