time derivative of involved position-derivative

Question

I am having a fair bit of trouble figuring out this differential. It's probably relatively straightforward but I can't wrap my head around it. I'm trying to work out:

$$ \frac{d}{dt}\left(-c^2\sqrt{1-\dot{x}^2\frac{1}{c^2}}\right) $$ Where c is just a constant and $\dot{x} = \frac{dx}{dt}$

Any help would be greatly appreciated!

Answer 1

Use the chain rule twice.

Let's consider the $1$-dimensional case first. Since $\frac{d}{d\dot{x}}(1-\dot{x}^2/c^2)=-2\dot{x}/c^2$,$$\frac{d}{d\dot{x}}[-c^2(1-\dot{x}^2/c^2)^{1/2}]=\frac{\dot{x}}{(1-\dot{x}^2/c^2)^{1/2}},$$so$$\frac{d}{dt}[-c^2(1-\dot{x}^2/c^2)^{1/2}]=\frac{\dot{x}\ddot{x}}{(1-\dot{x}^2/c^2)^{1/2}}.$$

With a bit of work, you can prove the vector generalization is$$\frac{d}{dt}[-c^2(1-\dot{x}^2/c^2)^{1/2}]=\frac{\dot{x}\cdot\ddot{x}}{(1-\dot{x}^2/c^2)^{1/2}}.$$

Answer 2

Simplify dividing by $c$ from the given $$ -c \frac{d}{dt}\sqrt{c^2- \dot x^{^2}}$$

The derivative of $\sqrt u$ is $\dfrac{1}{2\sqrt u},\text{so that is using Chain Rule }$

$$= -c \cdot \frac{-2 \dot x \ddot x}{2\sqrt{c^2- \dot x^{^2}}}$$

$$= \frac{c \dot x \ddot x}{\sqrt{c^2- \dot x^{^2}}}.$$

Answer 3

$y:=-c^2(1-(x')^2/c^2)^{1/2}$, and let $y\not =0;$

$y^2=c^4(1-(x')^2/c^2);$

Chain rule:

$2yy'=c^4(-2(x')(x'')/c^2);$

$y'=(-c^2)\dfrac{(x')(x'')}{(-c^2)(1-(x')^2/c^2)^{1/2}},$

$y'=\dfrac{(x')(x'')}{(1-(x')^2/c^2)^{1/2}}.$