We consider: $$\frac{d}{dt}\frac{1}{|\vec{x}(t)|}$$ where $\vec{x}(t)$ is a 3 dimensional vector.
I assumed I should apply the total derivative:
$$\frac{d}{dt}\frac{1}{|\vec{x}(t)|} = -x_1(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_1}{dt} -x_2(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_2}{dt} -x_3(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_3}{dt}$$
Is my assumption and computation correct?
Best Answer
The answer is correct and could be simplified as$$\frac{d}{dt}\frac{1}{|\vec{x}(t)|} =$$
$$ -x_1(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_1}{dt} -x_2(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_2}{dt} -x_3(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}}\frac{dx_3}{dt}=$$
$$-(x_1^2 + x_2^2 + x_3^2)^{-\frac{3}{2}} (x_1\frac{dx_1}{dt}+x_2\frac{dx_2}{dt}+x_3\frac{dx_3}{dt})=$$
$$- \frac{1}{|\vec{x}|^3} \left(\vec{x} \cdot \frac{d \vec{x}}{dt} \right).$$